MCQ Questions for Class 11 Engineering Maths Permutations And Combinations Quiz 11

The number of factors (excluding

$1$ and the expression itself) of the product of

$a^7b^4c^3def$ where

$a,b,c,d,e,f$ are all prime numbers is

0%
$1278$
0%
$1360$
0%
$1100$
0%
$1005$

Explanation

A factor of expression

$a^{7}b^{4}c^{3}def$ is simply the result of selecting one or more letters from

$7 a's, 4 b's, 3a's, d, e, f.$
The collection of letters can be observed as a collection of

$17$ objects out of which

$7$ are alike of one kind (a's),

$4$ are of second kind (b's),

$3$ are of third kind (c' s) and

$3$ are different

$(d, e, f)$
The number of selections =

$(1 + 7) (1 + 4) (1 + 3)(1+1)^{3} = 8 \times5\times 4\times 8 = 1280.$
But we have to exclude two cases : (i) When no letter is selected (ii) When all are selected.

So, the number of factors

$= 1280 -2 = 1278$
Hence, option

$'A'$ is correct.

$\space ^{n+1}C_{r+1}: ^nC_r : ^{n-1}C_{r-1} = 11:6:3$ , then find the values of

$n$ and

$r$ .

0%
$n=10,\space r=5$
0%
$n=9,\space r=4$
0%
$n=11,\space r=5$
0%
$n=10,\space r=4$

Explanation

Simplifying , we get

$6\:^{n+1}C_{r+1}=11\:^{n}C_{r}$

$\Rightarrow 6(n+1)=11(r+1)$

$\Rightarrow 6n-11r=5$ ... (i)
Also

$3\:^{n}C_{r}=6\:^{n-1}C_{r-1}$

$\Rightarrow 3n=6r$ ...(ii)
Substituting in 1, we get

$12r-11r=5$

$\Rightarrow r=5$
Hence

$n=10$

There are three papers of

$100$ marks each in an examination. In how many ways can a student get

$150$ marks such that he gets at least

$60%$ in two papers?

0%
$1480$
0%
$1488$
0%
$1520$
0%
$1526$

Explanation

Suppose the student gets at least

$60%$ marks in first two papers, then he just gets at most

$30$ marks in the third paper to make a total of

$150$ marks

So, the required numbers of ways = coefficient of

${ x }^{ 150 }$ in

${ \left( { x }^{ 60 }+{ x }^{ 61 }+...+{ x }^{ 100 } \right) }^{ 2 }\left( 1+x+{ x }^{ 2 }+...+{ x }^{ 30 } \right)$

$=$ coefficient of

${ x }^{ 30 }$ in

$\left( { \left( 1+x+...+{ { x }^{ 40 } } \right) }^{ 2 }\left( 1+x+...{ x }^{ 30 } \right) \right)$

$=$ coefficient of

${ x }^{ 30 }$ in

$\displaystyle { \left( \frac { 1-{ x }^{ 41 } }{ 1-x } \right) }^{ 2 }\left( \frac { 1-{ x }^{ 31 } }{ 1-x } \right)$

$=$ coefficient of

${ x }^{ 30 }$ in

${ \left( 1-x \right) }^{ -3 }$

$=^{ 30+3-1 }{ C }_{ 3-1 }=^{ 32 }{ C }_{ 2 }$

Thus the students get at least

$60%$ marks in the first two papers to get 150 marks as total in

$^{ 32 }{ C }_{ 2 }$ ways. But the two papers at least

$60%$ , can be chosen out of 3 papers in

$^{ 3 }{ C }_{ 2 }$ ways.

Hence,the required number of ways

$^{ 3 }{ C }_{ 2 }\times ^{ 32 }{ C }_{ 2 }=1488$

A person always prefers to eat parantha and vegetable dish in his meal. How many ways can he make his plate in a marriage party if there are three types of paranthas, four types of vegetable dishes, three types of salads, and two types of sauces?

0%
$3360$
0%
$4096$
0%
$3000$
0%
None of these

Explanation

The number of ways he can select at least one parantha is

$2^3-1=7$ .
The number of ways he can select at least one vegetable dish is

$2^4-1=15$ .

The number of ways he can select zero or more items from salads and sauces is

$2^5$ .
Hence, the total number of ways is

$7\times 15\times 32=3360$ .

Evaluate

$^{47}C_4 + \displaystyle\sum_{j=0}^{3}{^{50-j}C_3} + \sum_{k=0}^{5}{^{56-k}C_{53-k}}$

0%
$^{57}C_4$
0%
$^{57}C_5$
0%
$^{56}C_6$
0%
$^{56}C_5$

Explanation

Expanding , we get

$[\:^{47}C_{4}+\:^{47}C_{3}+\:^{48}C_{3}+\:^{49}C_{3}+\:^{50}C_{3}]+[\:^{56}C_{53}+\:^{55}C_{52}+\:^{54}C_{51}+\:^{53}C_{50}+\:^{52}C_{49}+\:^{51}C_{58}]$
Now

$\:^{n}C_{r}+\:^{n}C_{r+1}=\:^{n+1}C_{r+1}$
Hence

$[\:^{47}C_{4}+\:^{47}C_{3}+\:^{48}C_{3}+\:^{49}C_{3}+\:^{50}C_{3}$

$=\:^{48}C_{4}+\:^{48}C_{3}+\:^{49}C_{3}+\:^{50}C_{3}$
:
:

$=\:^{51}C_{4}$
Now

$\:^{56}C_{53}+\:^{55}C_{52}+\:^{54}C_{51}+\:^{53}C_{50}+\:^{52}C_{49}+\:^{51}C_{58}$

$=\:^{56}C_{3}+\:^{55}C_{3}+\:^{54}C_{3}+\:^{53}C_{3}+\:^{52}C_{3}+\:^{51}C_{3}$

$\:^{n}C_{r}=\:^{n}C_{n-r}$
Hence Adding we get

$[\:^{56}C_{3}+\:^{55}C_{3}+\:^{54}C_{3}+\:^{53}C_{3}+\:^{52}C_{3}+\:^{51}C_{3}]+\:^{51}C_{4}$

$=\:^{56}C_{3}+\:^{56}C_{4}$

$=\:^{57}C_{4}$

For

$2\le r \le n$ ,

$\left(\begin{matrix}n \\ r\end{matrix}\right) + 2\left(\begin{matrix}n \\ r-1 \end{matrix}\right) + \left(\begin{matrix}n \\ r - 2 \end{matrix}\right) \space =$

0%
$\left(\begin{matrix}n + 1 \\ r - 1\end{matrix}\right)$
0%
$2\left(\begin{matrix}n + 1 \\ r + 1\end{matrix}\right)$
0%
$2\left(\begin{matrix}n + 2 \\ r \end{matrix}\right)$
0%
$\left(\begin{matrix}n + 2 \\ r \end{matrix}\right)$

Explanation

$=^nC_r + 2. ^nC_{r-1} + ^nC_{r-2}$

$\displaystyle =^nC_r+\dfrac{2r}{(n-r+1)}.^nC_r+\dfrac{r(r-1)}{(n-r+1)(n-r+2)}.^nC_r$

$\displaystyle =^nC_r\left(\dfrac{n+r+1}{(n-r+1)}+\dfrac{r(r-1)}{(n-r+1)(n-r+2)}\right)$

$\displaystyle =^nC_r\left(\dfrac{n^{2}+3n+2}{(n-r+1)(n-r+2)}\right)$

$\displaystyle =^nC_r\left(\dfrac{(n+1)(n+2)}{(n-r+1)(n-r+2)}\right)$

$=^{n+2}C_r$

Hence, option 'D' is correct.

For any positive integer

$m,\space n \space(\mbox{ with } n\ge m) = ^nC_m$ .

$\left(\begin{matrix}n \\ m\end{matrix}\right) + \left(\begin{matrix}n - 1 \\ m\end{matrix}\right) + \left(\begin{matrix}n - 2 \\ m\end{matrix}\right) + ... + \left(\begin{matrix}m \\ m\end{matrix}\right) =$

0%
$\left(\begin{matrix}n+1 \\ m+1\end{matrix}\right)$
0%
$\left(\begin{matrix}n\\ m+1\end{matrix}\right)$
0%
$\left(\begin{matrix}n\\ m\end{matrix}\right)$
0%
$\left(\begin{matrix}n-1\\ m\end{matrix}\right)$

Explanation

$\begin{pmatrix} m \\ n \end{pmatrix}+\begin{pmatrix} m+1 \\ m \end{pmatrix}+\begin{pmatrix} m+2 \\ m \end{pmatrix}+...+\begin{pmatrix} n \\ m \end{pmatrix}$

$=\begin{pmatrix} m+1 \\ m+1 \end{pmatrix}+\begin{pmatrix} m+1 \\ m \end{pmatrix}+\begin{pmatrix} m+2 \\ m \end{pmatrix}+...+\begin{pmatrix} n \\ m \end{pmatrix}$

$=\begin{pmatrix} m+2 \\ m+1 \end{pmatrix}+\begin{pmatrix} m+2 \\ m \end{pmatrix}+...+\begin{pmatrix} n \\ m \end{pmatrix}$

$=\begin{pmatrix} m+3 \\ m+1 \end{pmatrix}+\begin{pmatrix} m+3 \\ m \end{pmatrix}+...+\begin{pmatrix} n \\ m \end{pmatrix}$
...................................

$=\begin{pmatrix} n \\ m+1 \end{pmatrix}+\begin{pmatrix} n \\ m \end{pmatrix}=\begin{pmatrix} n+1 \\ m+1 \end{pmatrix}$
Thus,

$\begin{pmatrix} n \\ m \end{pmatrix}+\begin{pmatrix} n-1 \\ m \end{pmatrix}+\begin{pmatrix} n-2 \\ m \end{pmatrix}+...+\begin{pmatrix} m \\ m \end{pmatrix}=\begin{pmatrix} n+1 \\ m+1 \end{pmatrix}$

The number of positive integers satisfying the inequality

$\quad ^{n+1}C_{n-2} - ^{n+1}C_{n-1} \le 100$ is

0%
One
0%
Eight
0%
Five
0%
None of these

Explanation

Given,

$\dfrac{(n+1)!}{3!(n-2)!}-\dfrac{(n+1)!}{2!.(n-1)!}$

$=\dfrac{(n-1)(n)(n+1)}{6}-\dfrac{n(n+1)}{2}$

$=(n)(n+1)\dfrac{n-1-3}{6}$

$(n)(n+1)\dfrac{n-4}{6}\leq100$

$\dfrac{n(n+1)(n-4)}{6}\leq 100$
Since

$n\epsilon N$
Hence

$2,3,4,5,6,7,8,9$ satisfy the above inequality.

The straight lines

$I_1, I_2, I_3$ are parallel and lie in the same plane. A total number of

$m$ points on

$I_1$ ;

$n$ points on

$I_2$ ;

$k$ points on

$I_3$ , the maximum number of triangles formed with vertices at these points are

0%
$^{m+n+k}C_3$
0%
$^{m+n+k}C_3 - ^mC_3 - ^nC_3 - ^kC_3$
0%
$^mC_3 + ^nC_3 + ^kC_3$
0%
None of these

Explanation

Toatl number of points are

$m+n+k$
the

$\Delta 's$ formed by these points

$=^{ m+n+k }{ { C }_{ 3 } }$
Points on the same line gives no triangle
Such

$\Delta 's$ are

$^{ m }{ { C }_{ 3 } }+^{ n }{ { C }_{ 3 } }+^{ k }{ { C }_{ 3 } }$
Therefore required number

$=^{ m+n+k }{ { C }_{ 3 } }-^{ m }{ { C }_{ 3 } }-^{ n }{ { C }_{ 3 } }-^{ k }{ { C }_{ 3 } }$

$r, s$ and

$t$ are prime numbers and

$p, q$ are positive integers such that the LCM of

$p,q$ is

$\displaystyle r^{2}t^{4}s^{2}$ then the number of ordered pair

$(p, q)$ is

0%
254
0%
252
0%
225
0%
224

Explanation

If 2 numbers A and B have LCM as L, where L =

${ a }^{ x }\times{ b }^{ y }\times{ c }^{ z }$ ., then, number of ordered pairs of numbers having the above LCM will be:

$(2x + 1) \times (2y + 1) \times (2z + 1)$

Let us consider r: we need the power r in either p or q to be at least

$2$ .

If the power of r in p is 2, then in q it should be 0 or 1 or 2

$\rightarrow3$ cases

If the power of r in q is 2, then in q it should be 0 or 1 or 2

$\rightarrow3$ cases

But (2, 2) has been counted twice. Thus, there are

$3 + 3 - 1 = 5$ cases

Total of five cases for the exponents of r such that we have the given LCM which is

$2\times 2 + 1$

Similarly, the others follow.
Here,

$x = 2, y = 4, z = 2.$

Therefore, the answer =

$(2\times2+1)\cdot (2\times4+1)\cdot (2\times2+1)=225$

Hence, (C) is correct.

$\displaystyle ^{7}C_{r} + 3 ^{7}C_{r+1} + 3 ^{7}C_{r+2} + ^{7}C_{r+3} > ^{10}C_{4}$ , then the quadratic equation whose roots are

$\displaystyle \alpha, \: \beta$ and

$\displaystyle \alpha^{r-1}, \: \beta^{r-1}$ have

0%
no common roots
0%
only one common root
0%
two common root
0%
none of these

Explanation

$\displaystyle ^{7}C_{r}+3^{7}C_{r+1}+3^{7}C_{r+2}+^{7}C_{r+3}> ^{10}C_{4}$

$\displaystyle \Rightarrow\>^{7}C_{r}+\>^{7}C_{r+1}+2\left (\>^{7}C_{r+1}+\>^{7}C_{r+2} \right )+\>^{7}C_{r+2}+\>^{7}C_{r+3}>\>^{10}C_{4}$

$\displaystyle \Rightarrow \>^{8}C_{r+1}+2\>^{8}C_{r+2}+\>^{8}C_{r+3}>\>^{10}C_{4}$

$\displaystyle \Rightarrow \>^{10}C_{r+3}>\>^{10}C_{4}$

$\Rightarrow r+3=5 \Rightarrow r=2$
Now roots of two different equation are

$\displaystyle \alpha ,\beta$ and

$\displaystyle \alpha^{r-1} ,\beta^{r-1}$

$\displaystyle$ i.e.

$\alpha ^{1},\beta ^{1}$ and

$\displaystyle \alpha ^{2-1},\beta ^{2-1}$ , i.e.

$\alpha ^{1},\beta ^{1}$ and

$\displaystyle \alpha ^{1},\beta ^{1}$

$\displaystyle \Rightarrow$ both roots are same
So Number of common roots are

$2$ .

In the figure,two 4-digit numbers are to be formed by filling the places with digits. The number of different ways in which the places can be filled by digits so that the sum of the numbers formed is also a 4-digit number and in no place the addition is with carrying, is

0%
$\displaystyle 55^4$
0%
220
0%
$\displaystyle 45^4$
0%
none of these

Explanation

The thousands place can be filled by 8 ways in both the cases whereas the other places can be filled by 9 ways in both the cases So, no of ways

$= 2.8.9.9.9 = 11664$ ways

$^nC_{r-1}=36, ^nC_r=84$ and

$^nC_{r+1}=126$ , then r is

0%
$1$
0%
$2$
0%
$3$
0%
none of these

Explanation

$^nC_{r-1}=36, ^nC_r=84$ ,

$^nC_{r+1}=126$

We know that

$\dfrac {^nC_{r-1}}{^nC_r}=\dfrac {r}{n-r+1}$

$\dfrac {36}{84}=\dfrac {r}{n-r+1}$

$\dfrac {r}{n-r+1}=\dfrac {3}{7}$

$3n-10r+3=0$ .....(1)

Also,

$\dfrac {^nC_r}{^nC_{r+1}}=\dfrac {r+1}{n-r}=\dfrac {84}{126}=\dfrac {2}{3}$

$2n-5r-3=0$ ....(2)

Solving (1) and (2), we get

$n=9$ and

$r=3$ .

The number of words of four letters containing equal number of vowels and consonants, repetition being allowed, is

0%
$\displaystyle 105^2$
0%
$\displaystyle 210 \: \times \: 243$
0%
$\displaystyle 105 \: \times \: 243$
0%
none of these

Explanation

Given, number of vowels =number of consonants
So, there will be 2 cases (i) no repetition(number of vowels =number of consonants=2)
(ii) both the vowel and consonant will be repeated twice(number of vowels =number of consonants=1)
The method that will be followed involves selection of 2 places out of the 4 places which can be done in 6 ways and then applying the above 2 cases
case (i): there will be 21 ways for secting the first consonant and 20 ways for the second and There will be 5 ways of selecting the first vowel and 4 ways of selecting the second.
that makes number of words in case (i) =

$6(21\times 20\times 5\times 4)$
case (ii): we have to select only 1 out of the 21 consonants and 5 vowels and both of them will be repeated twice

$\Longrightarrow$ number ofwords in case (ii) =

$6(21\times 5)$
\therefore number of words

$= 6(21\times 20\times 5\times 4)$ +

$6(21\times 5)$

$= 6(21\times 5)(20\times 4+1)$

$= 210\times243$

The value of

$\displaystyle ^{40}C_{31}+\sum_{j=0}^{10} \: ^{40+j}C_{10+j}$ is equal to

0%
$\displaystyle ^{51}C_{20}$
0%
$\displaystyle 2 . ^{50}C_{20}$
0%
$2 . ^{45}C_{15}$
0%
none of these

Explanation

Expanding, we get

We know that,

$^{n}C_{r-1}+^{n}C_{r}=^{n+1}C_{r}$

Therefore,

$\:^{40}C_{31}+[\:^{40}C_{10}+\:^{41}C_{11}+\:^{42}C_{12}+...\:^{50}C_{20}]$

$=\:^{40}C_{9}+[\:^{40}C_{10}+\:^{41}C_{11}+\:^{42}C_{12}+...\:^{50}C_{20}]$

$=(\:^{40}C_{9}+\:^{40}C_{10})+\:^{41}C_{11}+\:^{42}C_{12}+...\:^{50}C_{20}$

$=(\:^{41}C_{10}+\:^{41}C_{11})+\:^{42}C_{12}+...\:^{50}C_{20}$

$=\:^{42}C_{11}+\:^{42}C_{12}+...\:^{50}C_{20}$
:
:
:

$=\:^{49}C_{18}+\:^{49}C_{19}+\:^{50}C_{20}$

$=\:^{50}C_{19}+\:^{50}C_{20}$

$=\:^{51}C_{20}$

The number of different matrices that can be formed with elements 0,1,2 or 3, each matrix having 4 elements, is

0%
$\displaystyle 3\times 2^4$
0%
$\displaystyle 2 \times 4^4$
0%
$\displaystyle 3 \times 4^4$
0%
none of these

Explanation

${\textbf{Step 1: Consider a matrix which has 4 elements.}}$

${\text{As a matrix contains four elements so, the order of a matrix can be,}}$ $1 \times 4$ ${\text{or}}$ $4 \times 1$ ${\text{or}}$ $2 \times 2.$

${\text{Also in every such matrix, each element is independent and has 4 different choices}}$ $\left( {0, 1, 2, 3} \right)$

$\therefore$ ${\text{The number of ways to fill four places of a matrix of order 4}} \times {\text{1 by 0, 1, 2, 3}}$

$= {}^4{C_1} \times {}^4{C_1} \times {}^4{C_1} \times {}^4{C_1}$

$\therefore$ ${\text{The number of ways to fill four places of a matrix of order 4}} \times {\text{1 by 0, 1, 2, 3}}$ $= 4 \times 4 \times 4 \times 4$

$\left(\mathbf {\because {}^n{C_1} = n} \right)$

$\therefore$ ${\text{The number of ways to fill four places of a matrix of order 4}} \times {\text{1 by 0, 1, 2, 3}}$

$= {4^4} \ldots \left( 1 \right)$

${\textbf{Step 2: Find total number of matrices that can be formed.}}$

${\text{From equation }}\left( 1 \right)$ ${\text{we can say that, number of matrices in each order is }}{{\text{4}}^4}.$

${\text{Therefore,}}$

${\text{The number of matrices that can be formed}}$ $= {4^4} + {4^4} + {4^4}$

$\Rightarrow$ ${\text{The number of matrices that can be formed}}$ $= 3 \times {4^4}$

${\textbf{is correct answer.}}$

The value of

$\displaystyle \sum_{r=1}^{n}r(^{n}C_{r}+^{r}P_{r})$ is

0%
$\displaystyle n\cdot 2^{n-1}-1$
0%
$\displaystyle n\cdot 2^{n-1}+(n+1)!$
0%
$\displaystyle n\cdot 2^{n-1}+(n+1)!-1$
0%
$\displaystyle n^{2}+n+5$

Explanation

Expanding we get

$(1\:^{n}C_{1}+2\:^{n}C_{2}+...n\:^{n}C_{n})+(1\:^{1}P_{1}+2\:^{2}P_{2}+...n\:^{n}P_{n})$

$=\frac{d(1+x)^{n}}{dx}|_{x=1}+(1\:^{1}P_{1}+2\:^{2}P_{2}+...n\:^{n}P_{n})$

$=n(1+x)^{n-1}|_{x=1}+(1!+2(2!)+3(3!)+4(4!)+...n(n!))$

$=n2^{n-1}+(1!+2(2!)+3(3!)+4(4!)+...n(n!))$
Now consider

$S_{n}=1!+2(2!)+3(3!)+4(4!)+...n(n!)$

$S_{1}=1$

$=2!-1$

$S_{2}=5$

$=3!-1$

$S_{3}=1+4+18$

$=23$

$=4!-1$
Hence

$S_{n}=(n+1)!-1$
Hence the final answer is

$n2^{n-1}+(n+1)!-1$

The number of signals that can be given using any number of flags of 5 different colors, is

0%
$225$
0%
$325$
0%
$215$
0%
$315$

Explanation

Total number of signals can be made by using at a time one or more but not larger than five flags.
Now number of signals when r flags are used at a time from 5 flags is equal to the number of arrangement of 5 taking r at a time i.e

$^{5}P_{r}(r=1,2,...5)$

$\therefore$ Required ways

$^{5}P_{1}+^{5}P_{2}+....+^{5}P_{5}$

$=5+20+60+120+120$

$=325$

The value of the expression

$^{47}C_{4}+\sum_{f= 0}^{6}\ ^{52-f}C_{3}$ equals

0%
$^{47}C_{5}$
0%
$^{52}C_{5}$
0%
$^{52}C_{4}$
0%
$^{52}C_{3}$

Explanation

Simplifying, we get

$\:^{47}C_{4}+\:^{46}C_{3}+\:^{47}C_{3}+\:^{48}C_{3}+\:^{49}C_{3}+\:^{50}C_{3}+\:^{51}C_{3}+\:^{52}C_{3}$

$=\:^{46}C_{3}+\:^{47}C_{4}+\:^{47}C_{3}+\:^{48}C_{3}+\:^{49}C_{3}+\:^{50}C_{3}+\:^{51}C_{3}+\:^{52}C_{3}$

$=\:^{46}C_{3}+\:^{48}C_{4}+\:^{48}C_{3}+\:^{49}C_{3}+\:^{50}C_{3}+\:^{51}C_{3}+\:^{52}C_{3}$

$=\:^{46}C_{3}+\:^{49}C_{4}+\:^{49}C_{3}+\:^{50}C_{3}+\:^{51}C_{3}+\:^{52}C_{3}$
:
:
:

$=\:^{46}C_{3}+\:^{52}C_{4}+\:^{52}C_{3}$

$=\:^{46}C_{3}+\:^{53}C_{4}$

The exponent of 7 in the coefficient of the greatest term in the expansion of

$\displaystyle (1+x)^{200}$ is

0%
0
0%
1
0%
2
0%
3

Explanation

The greatest coefficient will be

$\:^{200}C_{100}$

$=\dfrac{200!}{100!.100!}$
Now the exponent of 7 in 200! will be

$[\dfrac{200}{7}]+[\dfrac{200}{49}]$

$=[28.57]+[4.08]$

$=28+4$

$=32$
The exponent of 7 in 100! will be

$[\dfrac{100}{7}]+[\dfrac{100}{49}]$

$=[14.28]+[2.04]$

$=14+2$

$=16$
Hence the power of 7 in

$(100!)^{2}$ will be

$16+16$

$=32$
Hence the powers of seven in the numerator and denominator will be equal.
Hence answer is

$0$ .

The mean value of

$^{20}C_0,\dfrac{^{20}C_2}{3},\dfrac{^{20}C_4}{5},\cdots ,\dfrac{^{20}C_{20}}{21}$ equals

0%
$\dfrac{2^{20}}{21}$
0%
$\dfrac{2^{19}}{21}$
0%
$\dfrac{2^{20}}{3\times 77}$
0%
$\dfrac{2^{19}}{33\times 7}$

Explanation

We know:

${ (1+x) }^{ n }=_{ 0 }^{ n }{ C }+_{ 1 }^{ n }{ C }x+_{ 2 }^{ n }{ C }{ x }^{ 2 }+...+_{ n }^{ n }{ C }{ x }^{ n }$
Integrating from 0 to 1:

$\dfrac { { (1+x) }^{ n+1 } }{ n+1 }-\dfrac { 1 }{ n+1 } =_{ 0 }^{ n }{ Cx+ }_{ 1 }^{ n }{ C\dfrac { { x }^{ 2 } }{ 2 } }+...+_{ n }^{ n }{ C\dfrac { { x }^{ n+1 } }{ n+1 } }$
Put

$x=1$ above:

$\dfrac { { 2 }^{ n+1 } }{ n+1 } =_{ 0 }^{ n }{ C+ }\dfrac { _{ 1 }^{ n }{ C } }{ 2 } +...+\dfrac { _{ n }^{ n }{ C } }{ n+1 }$
Put

$x=-1$ above:

$-\dfrac { 1 }{ n+1 } =-_{ 0 }^{ n }{ C+ }\dfrac { _{ 1 }^{ n }{ C } }{ 2 } -\dfrac { _{ 2 }^{ n }{ C } }{ 3 } ...+{ (-1) }^{ n }\dfrac { _{ n }^{ n }{ C } }{ n+1 }$
Subtracting the above 2 results and using

$n = 20$ :

$2(^{20}C_0+\dfrac { _{ 2 }^{ 20 }{ C } }{ 3 } +...+\dfrac { _{ 20 }^{ 20 }{ C } }{ 21 } )=\dfrac { { 2 }^{ 21 } }{ 21 }$
Hence, mean =

$\dfrac { \dfrac { 1 }{ 2 } \times \dfrac { { 2 }^{ 21 } }{ 21 } }{ 11 } =\dfrac { { 2 }^{ 20 } }{ 3\times77 }$ (since there are 11 terms)
Hence, (c) is correct.

$\displaystyle \frac{^{n}C_{r}+3^{n}C_{r+1}+3^{n}C_{r+2}+^{n}C_{r+3}}{^{n}C_{r}+4^{n}C_{r+1}+6^{n}C_{r+2}+4^{n}C_{r+3}+^{n}C_{r+4}}=\frac{r+k}{n+k}$ , then the value of

$k$ equals

0%
$1$
0%
$2$
0%
$4$
0%
None of these

Explanation

$\displaystyle \frac{^{n}C_{r}+3^{n}C_{r+1}+3^{n}C_{r+2}+^{n}C_{r+3}}{^{n}C_{r}+4^{n}C_{r+1}+6^{n}C_{r+2}+4^{n}C_{r+3}+^{n}C_{r+4}}$

$\displaystyle =\frac{r+k}{n+k}$

$\displaystyle \Rightarrow \frac{r+4}{n+4}=\frac{r+k}{n+k}$

$\Rightarrow k=4$

The expression

$^{n+4}C_{r}-^{n}C_{r}-3.^{n}C_{r-1}-3^{n}C_{r-2}-^{n}C_{r-3}$

0%
$^{n+3}C_{r-1}$
0%
$^{n+2}C_{r+1}$
0%
$^{n+4}C_{r+1}$
0%
$^{n+1}C_{r-1}$

Explanation

$^{n+4}C_{r}$ -

$^{n}C_{r}$ -3.

$^{n}C_{r-1}$ -3.

$^{n}C_{r-2}$ -

$^{n}C_{r-3}$

$\Longrightarrow$

$^{n+4}C_{r}$ -((

$^{n}C_{r}$ +

$^{n}C_{r-1}$ )+2.(

$^{n}C_{r-1}$ +

$^{n}C_{r-2}$ )+(

$^{n}C_{r-2}$ +

$^{n}C_{r-3})$

$\Longrightarrow$

$^{n+4}C_{r}$ -(

$^{n+1}C_{r}$ +2.

$^{n+1}C_{r-1}$ +

$^{n+1}C_{r-2}$ )

$\Longrightarrow$

$^{n+4}C_{r}$ -((

$^{n+1}C_{r}$ +

$^{n+1}C_{r-1}$ )+(

$^{n+1}C_{r-1}$ +

$^{n+1}C_{r-2}$ ))

$\Longrightarrow$

$^{n+4}C_{r}$ -(

$^{n+2}C_{r}$ +

$^{n+2}C_{r-1}$ )

$\Longrightarrow$

$^{n+3}C_{r}$ +

$^{n+3}C_{r-1}$ -

$^{n+3}C_{r}$

$\Longrightarrow$

$^{n+3}C_{r-1}$

$\displaystyle \sum _{ 0\le i\le }^{ }{ \sum _{ j\le 10 }^{ }{ ^{ 10 }{ { C }_{ j } }^{ j }{ { C }_{ i } } } }$ is equal to

0%
$3^{10}$
0%
$3^{10}-1$
0%
$2^{10}$
0%
$2^{10}-1$

Explanation

$\displaystyle \sum _{ 0\le i\le }^{ }{ \sum _{ j\le 10 }^{ }{ ^{ 10 }{ { C }_{ j } } } } \quad \quad ^{ j }{ { C }_{ i } }$

$\displaystyle ^{ 10 }{ { C }_{ 0 } }\times ^{ 0 }{ { C }_{ 0 } }\times ^{ 10 }{ { C }_{ 0 } }\left( ^{ 1 }{ { C }_{ 0 }\times ^{ 1 }{ { C }_{ 1 } } } \right) +^{ 10 }{ { C }_{ 2 } }\left( ^{ 2 }{ { C }_{ 0 } }+^{ 2 }{ { C }_{ 1 } }+^{ 2 }{ { C }_{ 2 } } \right) +^{ 10 }{ { C }_{ 3 } }\left( ^{ 3 }{ { C }_{ 0 } }+^{ 3 }{ { C }_{ 1 }+ }^{ 3 }{ { C }_{ 2 } }+^{ 3 }{ { C }_{ 3 } } \right)$

$\displaystyle +...+^{ 10 }{ { C }_{ 10 } }\left( ^{ 10 }{ { C }_{ 0 }+ }^{ 10 }{ { C }_{ 1 } }+...+^{ 10 }{ { C }_{ 0 } } \right)$

$\displaystyle =^{ 10 }{ { C }_{ 0 } }.{ 2 }^{ 0 }+^{ 10 }{ { C }_{ 1 } }.{ 2 }^{ 1 }+^{ 10 }{ { C }_{ 2 } }.{ 2 }^{ 2 }+...+^{ 10 }{ { C }_{ 10 } }.{ 2 }^{ 10 }$

$\displaystyle ={ \left( 1+2 \right) }^{ 10 }={ 3 }^{ 10 }.$

The sum

$\displaystyle \sum_{i=0}^{m}\binom{10}{i}\binom{20}{m-i}$ be maximum when m is

0%
15
0%
5
0%
10
0%
20

Explanation

Add upper suffixces if their sum is even the half of it will be the value r otherwise the value of r be sum of upper suffixces +1 divided by 2.
Given

$\displaystyle S=\sum_{t=0}^{m}\binom{10}{i}\binom{20}{m-i}=\binom{10}{0}\binom{20}{m}+\binom{10}{1}\binom{20}{m-1}+\binom{10}{2}\binom{20}{m-2}+...+\binom{10}{m}\binom{20}{0}$
In such cases we observe the upper suffixces and lower suffixces.
Upper suffices are 10 and 20 and lower suffixces taken the value (0...m).
In such cases we consider the two Binomial expansions

$\displaystyle \left ( 1+x \right )^{p}$ and

$\displaystyle \left ( 1+x \right )^{q}$ (Here p=10, q=20 or q=10, p=20) multiply them and collect the required exponent of x to which the given expression can be obtained.
Consider

$\displaystyle \left ( 1+x \right )^{10}\left ( 1+x \right )^{20}=\left ( ^{10}C_{0}+^{10}C_{2}x^{2}+...+^{10}C_{10^{x^{10}}} \right )\left ( ^{20}C_{0} +^{20}C_{1}x+...+^{20}C_{20^{x^{20}}}\right )$

$\displaystyle \left ( 1+x \right )^{30}=\left ( ^{10}C_{0} +^{10}C_{1^{x^{1}}}+...+^{10}C_{10^{x^{10}}}\right ).\left ( ^{20}C_{0}+^{20}C_{1}x^{1}+...+^{20}C_{m^{x^{m}}}+...+^{20}C_{20^{x^{20}}} \right )$
and collecting coefficient of

$\displaystyle x^{m}$ both side, we get

$\displaystyle ^{30}C_{m}=\binom{10}{0}\binom{20}{m}+\binom{10}{1}\binom{20}{m-2}+...+\binom{10}{m}\binom{20}{0}$ now for max value of the expression will be at

$\displaystyle m=\frac{30}{2}=15$ as upper suffix is even

$\displaystyle \therefore m=15$ which is given in (a).

The value of the expression

$\displaystyle2^{k}\binom{n}{0}\binom{n}{k}-2^{k-1}\binom{n}{1}\binom{n-1}{k-1}+2^{k-2}\binom{n}{2}\binom{n-2}{k-2}..+(-1)^{k}\binom{n}{k}\binom{n-k}{0}$ is

0%
$\displaystyle \binom{n}{k}$
0%
$\displaystyle \binom{n+1}{k}$
0%
$\displaystyle \binom{n+1}{k+1}$
0%
$\displaystyle \binom{n-1}{k-1}$

Explanation

$\displaystyle 2^{k}\binom{n}{0}\binom{n}{k}-2^{k-1}\binom{n}{1}\binom{n-1}{k-1}+2^{k-2}\binom{n}{2}\binom{n-2}{k-2}..+(-1)^{k}\binom{n}{k}\binom{n-k}{0}$
We know that

$\displaystyle\binom{n}{0}\binom{n}{k}=\binom{k}{0}\binom{n}{k}$

$\displaystyle\binom{n}{1}\binom{n-1}{k-1}=\displaystyle\frac { n! }{ (n-1)! } \frac { (n-1)! }{ (n-k)!(k-1)! } =\frac { k(n!) }{ k!(n-k)! } =\binom{k}{1}\binom{n}{k}$

In the similar manner

$\displaystyle\binom{n}{k}\binom{n-k}{0}=\binom{k}{k}\binom{n}{k}$

$\Rightarrow \displaystyle 2^{k}\binom{k}{0}\binom{n}{k}-2^{k-1}\binom{k}{1}\binom{n}{k}+2^{k-2}\binom{k}{2}\binom{n}{k}..+(-1)^{k}\binom{k}{k}\binom{n}{k}$

$\Rightarrow \displaystyle \binom{n}{k}\left[2^{k}\binom{k}{0}-2^{k-1}\binom{k}{1}+2^{k-2}\binom{k}{2}...+(-1)^{k}\binom{k}{k}\right]$

$\Rightarrow \displaystyle \binom{n}{k}\left(2-1\right)^{k}$

$=\displaystyle \binom{n}{k}$
Hence, option A.

$\displaystyle ^{100}C_{3} = 161700$ , then

$^{100}C_{97}$ is equal to___.

0%
53,900
0%
40,425
0%
1,61,700
0%
16,17,000

Explanation

A properties of combinations is

$^nC_r = ^nC_{r-1}$

So,

$^{100}C_{97} = ^{100}C_{ 100-97 } = ^{100}C_{ 3 } = 1,61,700$

A road network as shown in the figure connect four cities. In how many ways can you start from any city (say A) and come back to it without travelling on the same road more than once ?

0%
8
0%
12
0%
9
0%
16

Explanation

Paths are shown as :

Similarly if we start from A towards B we get another $4$ paths.

Similarly if we start from A towards B again $3$ paths.

Total different paths $4\times3=12$

$\displaystyle ^{n}C_{3}= ^{n}C_{5'}$ then find the value of n:

0%
$9$
0%
$10$
0%
$8$
0%
$7$

Explanation

$^nC_r = \dfrac { n! }{ (r!)(n-r)! }$

Given,

$^nC_3 =^nC_5$

$=> \dfrac { n! }{ (3!) (n-3)! } =\dfrac { n! }{ 5! (n -5)! }$

$=> 3 ! \times (n - 3) \times (n - 4) \times (n-5) ! = 5 \times 4 \times 3 ! \times (n-5)!$

$=> (n - 3) \times (n - 4) = 5 \times 4$

From this,

$n - 3 = 5$

$=> n = 8$

${ _{ }^{ n }{ C } }_{ 4 },{ _{ }^{ n }{ C } }_{ 5 }$ and

${ _{ }^{ n }{ C } }_{ 6 }$ are in AP, then

$n$ is

0%
$7$ or $14$
0%
$7$
0%
$14$
0%
$14$ or $21$

Explanation

$\textbf{Step 1: Find the value of n}$

$\text{If a,b,c are in A.P,then}$

$2b=a+c$

$\text{So applying the same conditions, we have}$

$\Rightarrow$

$2 \times\left({ }^{n} c_{5}\right)=\left({ }^{n} C_{6}\right)+\left({ }^{n} C_{4}\right)$

$\text{We know that,}$

${ }^{n} C_{r} = \dfrac{ n !}{r !(n-r) !}$

$\Rightarrow$

$\dfrac{2 \times n !}{5 !(n-5) !}=\dfrac{n !}{6 !(n-6) !}+\dfrac{n !}{4 !(n-4) !}$

$\Rightarrow \dfrac{2}{5 !(n-5) !}=\dfrac{1}{6(n-6) !}+\dfrac{1}{4 !(n-4) !}$

$\Rightarrow \dfrac{2}{5 !(n-5)(n-6) !}=\dfrac{1}{6 !(n-6) !}+\dfrac{1}{4 !(n-4)(n-5)(n-6)!}$

$\Rightarrow \dfrac{2}{5 !(n-5)}=\dfrac{1}{6 !}+\dfrac{1}{4 !(n-4)(n-5)}$

$\Rightarrow \dfrac{2}{5 \times 4 ! \times(n-5)}=\dfrac{1}{6 \times 5 \times 4 !}+\dfrac{1}{4 !(n-4)(n-5)}$

$=\dfrac{2}{5 \times(n-5)}=\dfrac{1}{6 \times 5}+\dfrac{1}{(n-4)(n-5)}$

$\Rightarrow \dfrac{2}{5 \times(n-5)}-\dfrac{1}{(n-4)(n-5)}=\dfrac{1}{6 \times 5}$

$\Rightarrow 12 n-78=n^{2}-9 n+20$

$\Rightarrow n^2-21n+98=0$

$\Rightarrow (n-7)(n-14)=0$

$\therefore n=14$

$\text{(or)}$

$n=7$

$\textbf{Hence, option A is correct.}$

$^{19}C_r$ and

$^{19}C_{r-1}$ are in the ratio 2:3, then find

$^{14}C_r$

0%
91
0%
81
0%
71
0%
61

Explanation

We know that,

${ { n }_{ C } }_{ r }=\dfrac { n! }{ r!(n-r)! }$

Given,

$\dfrac { { { 19 }_{ C } }_{ r } }{ { { 19 }_{ C } }_{ r-1 } } =\quad \dfrac { 2 }{ 3 }$

$=>\dfrac { \dfrac { 19! }{ r!(19-r)! } }{ \dfrac { 19! }{ (r-1)!(19-r+1)! } } =\quad \dfrac { 2 }{ 3 }$

$=> \dfrac { (r-1)!(20-r)! }{ r!(19-r)! } =\quad \dfrac { 2 }{ 3 }$

$=\dfrac { (r-1)!\times (20-r) \times (19-r)! }{ r\quad \times (r-1)!\times (19-r)! } =\quad \dfrac { 2 }{ 3 }$

$=> \dfrac { 20-r }{ r\quad } =\quad \dfrac { 2 }{ 3 }$

$=> r = 12$

So,

${ { 14 }_{ C } }_{ 12 }=\dfrac { 14! }{ 12!(14-12)! } =\dfrac { 14! }{ 12!(2)! } =\dfrac { 14\times 13\times 12! }{ 12!\times 2 } =\quad 7\times 13=91$

A bag contains n white and n black balls. Pairs of balls are drawn at random without replacement successively, until the bag is empty. If the number of ways in which each pair consists of one white and one black ball is 14,then n =

0%
6
0%
5
0%
4
0%
3

Explanation

According to the given condition:

$(^nC_1\ ^nC_1)$

$(^{n-1}C_1\ ^{n-1}C_1)..........$

$(^1C_1\ ^1C_1)=14400$

$\Rightarrow (^nC^{n-1}C_1.......\ ^1C_1)^2=(120)^2$

$\Rightarrow n(n-1)(n-2)....\ 1=120$

$\Rightarrow n!=5!$

$\Rightarrow n=5$

Three persons entered a railway compartment in which

$5$ seats were vacant. Find the number of ways in which they can be seated

0%
$30$
0%
$45$
0%
$120$
0%
$60$

Explanation

The first person can choose to sit in any of the

$5$ seats in

$5$ ways.

Then the second person can choose to sit in any of the remaining

$4$ seats in

$4$ ways.

Lastly, the third person can choose to sit in any of the remaining

$3$ seats in

$3$ ways.

So, total number of ways

$= 5 \times 4 \times 3 = 60$ ways

How many

$3$ digit numbers can we make using the digits

$4,5,7$ and

$9$ and where repetition is allowed?

0%
${3}^{4}$
0%
${3}^{3}$
0%
${4}^{4}$
0%
$12$
0%
${4}^{3}$

Explanation

We have four digits.

The numbers we need to make have

$3$ digits.

There are

$4$ choices for the first digit,

$4$ choices for the second digit and

$4$ choices for the third digit since repetition is allowed.

Hence, the total number of

$3$ digit numbers that can be made is equal to:

$4\times 4\times 4={ 4 }^{ 3 }$

Hence, option E is correct.

A password for a computer system requires exactly

$6$ characters. Each character can be either one of the

$26$ letters from A to Z or one of the ten digits from

$0$ to

$9$ . The first character must be a letter and the last character must be a digit. How many different possible passwords are there?

0%
Less than $10^{7}$
0%
Between $10^{7}$ and $10^{8}$
0%
Between $10^{8}$ and $10^{9}$
0%
Between $10^{9}$ and $10^{10}$
0%
More than $10^{10}$

Explanation

The password required has to be of 6 characters and the first of those should be one of the 26 letters, the last must be one of the 10 digits and the remaining four can be either of those 36.

Thus, the possibilities for first place become 26, for the next four places become 36, and for the last place become 10.

The total number of possibilities therefore become

$36 \times (26)^4 \times 10 = 16,45,11,360$

This can be written as

$1.6451136 \times 10^8$ and thus lies between

$10^8$ and

$10^9$

The exponent of

$5$ in

$^{120}C_{60}$ , is

0%
$1$
0%
$0$
0%
$2$
0%
$3$

Explanation

Exponent of

$5$ in

$^{ 120 }{ C }_{ 60 }$

$={}^{ n }{ C }_{ r }=\dfrac { n! }{ (n-r)!r! } \\ \\ ={}^{ 120 }{ C }_{ 60 }=\dfrac { 120! }{ 60!\times 60! } \\ =\dfrac { 120\times 119\times 118\times .....61 }{ 60\times 59\times 58\times .....2\times 1 }$

Number of multiples of

$5$ in Numerator

$=$ Denominator

$=14$

Overall exponent of

$5=0$

$^nC_{r-1}=36, ^nC_r = 84$ and

$^nC_{r+1} = 126$ then the value of

$^nC_8$ is:

0%
$10$
0%
$7$
0%
$9$
0%
$8$

Explanation

$^nC_{r - 1} = \cfrac{n!}{(r - 1)! (n - r + 1)!} \ \ ..(1)$

$^nC_r = \cfrac{n!}{r!(n - r)!} \ \ ..(2)$

$^nC_{r + 1} = \cfrac{n!}{(r + 1)!(n - r - 1)!} \ \ ..(3)$

Dividing equation

$(2)$ by

$(1)$ , we get

$\cfrac{84}{36} = \cfrac{n - r + 1}{r}$

$\therefore 7r = 3n - 3r + 3$

$\therefore 10r = 3n + 3 \ \ ..(4)$

Dividing equation

$(3)$ by

$(2)$ , we get

$\cfrac{126}{84} = \cfrac{n - r}{r + 1}$

$\therefore 3r + 3 = 2n - 2r$

$\therefore 2n = 5r + 3 \ \ ..(5)$

Multiplying equation

$(5)$ by

$2$ and adding that to equation

$(4)$ , we get

$4n + 10r = 3n + 3 + 10r + 6$

$\therefore n = 9$

$\therefore \displaystyle ^nC_{8} = \ ^{9}C_{8} = 9$

The value of the expression

$^{47}C_4$ +

$\sum_{j=1}^{5} { }^{52-j}C_3$ is

0%
$^{51}C_4$
0%
$^{52}C_4$
0%
$^{52}C_3$
0%
$^{53}C_4$

Explanation

$^{ 47 }{ C }_{ 4 }+\sum _{ j=1 }^{ 5 }{ ^{ 52-j }{ C }_{ 3 } } =?\\ ^{ 47 }{ C }_{ 4 }+^{ 51 }{ C }_{ 3 }+^{ 50 }{ C }_{ 3 }+^{ 49 }{ C }_{ 3 }+^{ 48 }{ C }_{ 3 }+^{ 47 }{ C }_{ 3 }$

We Know ,

$^{ n }{ C }_{ r }+^{ n }{ C }_{ n-r }=^{ n+1 }{ C }_{ r }\\ \Rightarrow ^{ 47 }{ C }_{ 4 }+^{ 47 }{ C }_{ 3 }+^{ 48 }{ C }_{ 3 }+^{ 49 }{ C }_{ 3 }+^{ 50 }{ C }_{ 3 }+^{ 51 }{ C }_{ 3 }\\ \Rightarrow ^{ 48 }{ C }_{ 4 }+^{ 48 }{ C }_{ 3 }+^{ 49 }{ C }_{ 3 }+^{ 50 }{ C }_{ 3 }+^{ 51 }{ C }_{ 3 }\\ \Rightarrow ^{ 49 }{ C }_{ 4 }+^{ 49 }{ C }_{ 3 }+^{ 50 }{ C }_{ 3 }+^{ 51 }{ C }_{ 3 }\\ \Rightarrow ^{ 50 }{ C }_{ 4 }+^{ 50 }{ C }_{ 3 }+^{ 51 }{ C }_{ 3 }\\ \Rightarrow ^{ 51 }{ C }_{ 4 }+^{ 51 }{ C }_{ 3 }\\ \Rightarrow ^{ 52 }{ C }_{ 4 }$

Hence the correct answer is

$^{ 52 }{ C }_{ 4 }$

$^{n-1}C_r = (k^2 - 3) ^nC_{r+1} ,$ then k belongs to the interval

0%
$[\sqrt{-3}, \sqrt3 ]$
0%
$(-\infty,-2)$
0%
$(2, \infty)$
0%
$(\sqrt3, 2]$

Explanation

$^{ n-1 }{ C }_{ r }=({ K }^{ 2 }-3)^{ n }{ C }_{ r+1 }\\ \Rightarrow ^{ n }{ C }_{ r }=\cfrac { n! }{ (n-r)!r! } \\ \Rightarrow \cfrac { \cfrac { (n-1)! }{ (n-r-1)!r! } }{ \cfrac { n! }{ (n-r-1)!(r+1)! } } ={ K }^{ 2 }-3\\ \Rightarrow \cfrac { r }{ n } ={ K }^{ 2 }-3$

We know in the value of

$^{ n }{ C }_{ r }$

value of

$n\ge r$

$\Rightarrow \cfrac { r }{ n }$ value lies between

$[0,1]$

$\Rightarrow { K }^{ 2 }-3\in \left[ 0,1 \right] \\ \Rightarrow { K }^{ 2 }\in \left[ 3,4 \right] \\ \Rightarrow { K }\in \left[ \sqrt { 3 } ,2 \right]$

$^{20}C_r = ^{20}C_{r-10}$ , then the value of

$^{18}C_r$ is:

0%
$4896$
0%
$5432$
0%
$816$
0%
$1632$

Explanation

Given

$^{ 20 }{ C }_{ r }=^{ 20 }{ C }_{ r-10 }$

We Know

$^{ n }{ C }_{ r }=^{ n }{ C }_{ n-r }\\ \Rightarrow r+(n-r)=n\\ \Rightarrow r+r-10=20\\ \Rightarrow 2r=30\\ \Rightarrow r=15\\ \Rightarrow ^{ 18 }{ C }_{ r }=^{ 18 }{ C }_{ 15 }\\ \Rightarrow \dfrac { 18! }{ 15!\times 3! } \\ \Rightarrow \dfrac { 18\times 17\times 16 }{ 3\times 2\times 1 } \\ \Rightarrow 48\times 17\\ \Rightarrow 816$

Hence the correct answer is

$816$

The exponent of

$7$ in

$^{120}C_{50}$ , is

0%
$0$
0%
$2$
0%
$4$
0%
None of these

Explanation

Exponent of $7$ in $^{ 120 }{ C }_{ 50 }$

$^{ n }{ C }_{ r }=\dfrac { n! }{ (n-r)!r! } \\ \\ \Rightarrow ^{ 120 }{ C }_{ 50 }=\dfrac { 120! }{ 70!\times 50! } \\ =\dfrac { 120\times 119\times 118\times .....71 }{ 50\times 49\times 48\times .....2\times 1 }$

Number of multiples of $7$ in Numerator $=$ Denominator $=8$

Therefore Exponent of $7$ is $0$

For a chess tournament

$13$ people were selected for quarter finals. Each person plays two matches with the other. How many matches have been held in the whole tournament?

0%
$144$
0%
$156$
0%
$185$
0%
$116$

Explanation

$13$ People Each people plays two matches

Selecting two players and letting them play

$={}^{ 13 }{ P }_{ 2 }$

$=\dfrac { 13\times 12\times 11\times 10\times .......3\times 2\times 1 }{ 11! }$

$=13\times 12$

$=156$

Hence Number of matches is

$156$

The area of regular polygon of n sides with length of side

$\sqrt{3}$ , where n > 1 and satisfies the relation

$\displaystyle \sum_{r=0}^{n} \frac{n^2-3n+3}{2\ ^nC_r}=\sum_{r=0}^n\frac{r}{^nC_r}$ is :

0%
$30\sqrt{3}$
0%
$\dfrac{3\sqrt{3}}{4}$
0%
$\dfrac{5\sqrt{3}}{4}$
0%
$\dfrac{3\sqrt{3}}{2}$

Explanation

$\displaystyle \sum\dfrac r{^nC_r} = \sum\dfrac{n-r}{^nC_r} = \dfrac n2\sum\dfrac1{^nC_r}$

$\dfrac{n^2-3n+3}{2} \sum\dfrac1{^nC_r} \Rightarrow n^2-3n+3=n$

$n^2-4n+3=0 \Rightarrow n=3 (\because n>1)$

$\therefore$ Area of the equilateral triangle with side

$\sqrt3$ is

$\dfrac{3\sqrt3}4$

$12$ people came for a carroms tournament. All were divided into pairs of

$2$ each. After all the matches if it found that half of the selected pairs had to play three matches to decide the winner in a best of three process, how many matches have been held?

0%
$165$
0%
$185$
0%
$198$
0%
$132$

$^{19}C_{3r} = ^{19}C_{r+3}$ , then r is equal to:

0%
$5$
0%
$4$
0%
$3$
0%
$2$

Explanation

$^{ 19 }{ C }_{ 3r }=^{ 19 }{ C }_{ r+3 }$

We Know

$^{ n }{ C }_{ r }=^{ n }{ C }_{ n-r }\\ \Rightarrow r+(n-r)=n\\ \Rightarrow 3r+r+3=19\\ \Rightarrow 4r=16\\ \Rightarrow r=4$

Therefore The value of

$r$ is

$4$

How many alphabets need to be there in a language if one were to make

$1$ million distinct

$3$ digit initials using the alphabets of the language?

0%
$10$
0%
$100$
0%
$56$
0%
$26$

Explanation

Digits with initials

$=3$

Total number of initials

$=1million=1000000$

Let the alphabets in language be

$x$

initials

$={ x }^{ 3 }=1000000={ 10 }^{ 6 }\\ \Rightarrow x={ 10 }^{ 2 }$

Therefore

$100$ alphabets are present

After every get-together every person present shakes the hand of every other person. If there were 105 handshakes in all, how many persons were present in the party?

0%
$16$
0%
$15$
0%
$13$
0%
$14$

Explanation

Total number of handshakes

$=105$

Let number of person be

$=x$

Number of handshakes

$=\cfrac { x(x-1) }{ 2 } =105\\ \Rightarrow x(x-1)=210\\ x=15$

Hence the correct answer is

$15$

$8064$ is resolved into all possible product of two factors. Find the number of ways in which this can be done?

0%
$24$
0%
$21$
0%
$20$
0%
None of these

Explanation

$8064=2^{ 7 }\times { 3 }^{ 2 }\times 7$

Resolution of a number into factors (and thier products).

depends on powers of primes.

Number of possible products are

$=\left( 7+1 \right) \left( 2+1 \right) \left( 1 \right) \\ =8\times 3\times 1\\ =24$

How many straight lines can be formed from

$11$ points in a plane out of which no three points are collinear?

0%
$66$
0%
$50$
0%
$55$
0%
$60$

Explanation

Number of points

$=11$

No three are collinear

For a straight lines Two end points are needed

Number of straight lines needed

$\Rightarrow ^{ 11 }{ C }_{ 2 }\\ \Rightarrow \cfrac { 11! }{ 2!\times 9! } \\ \Rightarrow \cfrac { 11\times 10 }{ 2 } \\ \Rightarrow 55 lines$

Option

$: C$

A college offers

$7$ courses in the morning and

$5$ in the evening. Find possible number of choices with the student who want to study one course in the morning and one in the evening.

0%
$35$
0%
$12$
0%
$49$
0%
$25$

Explanation

Number of courses in the morning

$=7$

Selecting one of the

$7=^{ 7 }{ C }_{ 1 }$ Ways

$=7$ ways

Number of courses in the evening

$= 5$

Selecting one of the

$5=^{ 5 }{ C }_{ 1 }$ Ways

$= 5$ ways

Total number of ways

$=7\times 5$ Ways

$=35$ Ways

CBSE Questions for Class 11 Engineering Maths Permutations And Combinations Quiz 11 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Maths Permutations And Combinations Quiz 11 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

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