If $$\left( 2\le r\le n \right) $$, then $${ _{ }^{ n }{ C } }_{ r }+2{ _{ }^{ n }{ C } }_{ r+1 }+{ _{ }^{ n }{ C } }_{ r+2 }$$ is equal to

$$2.{ _{ }^{ n+2 }{ C } }_{ r+2 }\quad $$

$${ _{ }^{ n+r }{ C } }_{ r+1 }$$

MCQ Questions for Class 11 Engineering Maths Permutations And Combinations Quiz 7

$n$ is an integer greater than

$1$ , then

$a-{ _{ }^{ n }{ C } }_{ 1 }(a-1)+{ _{ }^{ n }{ C } }_{ 2 }(a-2)+....+{ (-1) }^{ n }(a-n)$ is equal to

0%
$a$
0%
$0$
0%
${a}^{2}$
0%
${2}^{n}$

Explanation

$a-{ _{ }^{ n }{ C } }_{ 1 }(a-1)+{ _{ }^{ n }{ C } }_{ 2 }(a-2)+....+{ (-1) }^{ n }(a-n)$

$=a-{}^{ n }{ C }_{ 1 }a+{}^{n}C_{1}+{}^{ n }{ C }_{ 2 }a-2\ {}^{n}C_{2}+....+(-1)^{n}a-(-1)^{n}n\ {}^{n}C_{n}$

$=a\left\{ { C }_{ 0 }-{ C }_{ 1 }+{ C }_{ 2 }-{ C }_{ 3 }+......{ (-1) }^{ n }{ C }_{ n } \right\} +\left\{ { C }_{ 1 }-{ 2C }_{ 2 }+3{ C }_{ 3 }-......{ (-1) }^{ n }{ n.C }_{ n } \right\}$

$=a.0+0=0$

Choose

$3, 4, 5$ points other than vertices respectively on the sides

$AB, BC$ and

$CA$ of a

$\triangle ABC$ . The number of triangles that can be formed by using only these points as vertices, is

0%
$220$
0%
$217$
0%
$215$
0%
$210$
0%
$205$

Explanation

Required number of triangles that can be formed by using only given points as vertices

$= ^{12}C_{3} - \left \{^{3}C_{3} + ^{4}C_{3} + ^{5}C_{3}\right \}$

$= \dfrac {12\times 11\times 10}{3\times 2\times 1} - \left \{1 + 4 + \dfrac {5\times 4}{2\times 1}\right \}$

$= 220 - (5 + 10)$

$= 220 - 15 = 205$

Total number of words that can be formed using all letters of the word

$\text{BRIJESH}$ that neither begins with

$I$ nor ends with

$B$ is equal to.

0%
$4920$
0%
$3720$
0%
$3600$
0%
$4800$

Explanation

Total number of words formed without any restriction

$=7!$

Total number of words beginning with

$I=6!$

Total number of words beginning with

$B=6!$

Total number of words beginning with I and ends with

$B=5!$

Therefore, required number of words

$=7!-6!-6!+5!$

$=5040-2\times 720+120=3720$

$^{n}C_{15} = ^{n}C_{8}$ , then the value of

$^{n}C_{21}$ is

0%
$254$
0%
$250$
0%
$253$
0%
None of these

Explanation

Given,

${}^{n}C_{15} = {}^{n}C_{8}$

$\Rightarrow {}^{n}C_{n - 15} = {}^{n}C_{8}$

$\therefore n - 15 = 8\Rightarrow n = 23$
Now,

${}^{n}C_{21} = {}^{23}C_{21}$

$= \dfrac {(23)!}{2! \times (21)!}$

$= \dfrac {23\times 22}{2} = 23\times 11 = 253$ .

What was the bus charge per student?

0%
Rs 194
0%
Rs 184
0%
Rs 187
0%
Rs 199

A village has

$10$ players. A team of

$6$ players is to be formed.

$5$ members are chosen first out of these

$10$ players and then the captain from the remaining players. Then the total number of ways of choosing such teams is

0%
$1260$
0%
$210$
0%
$({ _{ }^{ 10 }{ C } }_{ 6 })5!$
0%
${ _{ }^{ 10 }{ C } }_{ 5 }6$

Explanation

A village has

$10$ players.

Now a team of

$6$ players is to be formed.

Thus

$1^{st}$ we select the

$5$ players from total

$10$ players in

${ _{ }^{ 10 }{ C } }_{ 5 }$ ways and

$6^{th}$ person will select from the remaining

$5$ players in

$5$ ways.
Total number of ways

$={ _{ }^{ 10 }{ C } }_{ 5 }\times 5=1260$

An alphabet contains a

$A^{'s}$ and b

$B^{'s}$ . (In all a+b letters ). The number of words each containing all the

$A^{'s}$ and any number of

$B^{'s}$ , is

0%
$^{a+b}C_b$
0%
$^{a+b+1}C_a$
0%
$^{a+b+1}C_b$
0%
none of these

The maximum number of intersection points of n circles and n straight lines , among themselves is 80.The value of n is

0%
$7$
0%
$6$
0%
$5$
0%
$8$

The number of ways of distributing

$8$ identical balls in

$3$ distinct boxes, so that none of the boxes is empty is

0%
$5$
0%
$21$
0%
${ 3 }^{ 8 }$
0%
$^{ 8 }{ { C }_{ 3 } }$

Explanation

The total number of ways of dividing

$8$ identical balls in

$3$ distinct boxes, so that one of the boxes is empty, is

$^{ 8-1 }{ { C }_{ 3-1 } }$ .

That is further equal to

$^{ 7 }{ { C }_{ 2 } }=\dfrac { 7! }{ 2!5! } =21$ .

$p> q$ , the number of ways of

$p$ men and

$q$ women can be seated in a row so that no two women sit together is

0%
$\cfrac { p!q! }{ \left( p+q \right) ! }$
0%
$\cfrac { \left( p+q \right) ! }{ p!\left( q+! \right) ! }$
0%
$\cfrac { p!\left( q+1 \right) ! }{ \left( p-q+1 \right) ! }$
0%
$\cfrac { p!(p+1)! }{ \left( p-q+1 \right) ! }$

Explanation

Arrangement of

$p$ men in a row

$=p!$

$\therefore$ Required number of ways

$={ p! }^{ p+1 }{ P }_{ q }=\cfrac { p!(p+1)! }{ \left( p-q+1 \right) ! }$

Total number of arrangements of the letters of the word SUCCESS such that both

$C's$ are together and no two

$S's$ are together is

0%
$12$
0%
$24$
0%
$96$
0%
$120$

Explanation

$CC$ are together and no two

$'S'$ are together

$\therefore$ Consider

$CC$ as single object.

$U, CC, E$ can be arranged in

$3!$ ways

$\times U\times CC \times E \times$
Now three

$'S'$ are to the placed in four available place as

$^{4}c_{3}$ ways

$\therefore$ Required number of ways are

$3! \times ^{4}C_{3} = 24$ .
Hence choice (b) is correct.

$\alpha ,\beta , \gamma$ are three consecutive integers. If these integers are raised to first, second and third positive powers respectively, and added then they form a perfect square, the square root of which is equal to the sum of these integers. Also,

$\alpha < \beta < \gamma$ . Then,

$\gamma$ is equals to:

0%
$3$
0%
$14$
0%
$5$
0%
$11$

Explanation

Let the numbers be

$n-1,n,n+1$

As per the given information, we have

$(n-1)+{ n }^{ 2 }+{ (n+1) }^{ 3 }={ (n-1+n+n+1) }^{ 2 }$

$\Rightarrow { n }^{ 3 }-5{ n }^{ 2 }+4n=0$

$\Rightarrow n=0,1,4$

For

$n=0,$ the numbers are:

$-1,0,1$ this is out as all the numbers should be positive

For

$n=1,$ the numbers are:

$0,1,2$ this is out as all the numbers should be positive (‘0’ can’t be taken as positive)

For

$n=4,$ the numbers are:

$3,4,5$

We have got largest number which is

$5$

Hence, the value of

$\gamma$ is

$5$ .

Out of

$15$ persons

$10$ can speak Hindi and

$8$ can speak English. If two persons are chosen at random, then the probability that one person speaks Hindi only and the other speaks both Hindi and English is

0%
$5/3$
0%
$7/12$
0%
$1/5$
0%
$2/5$

Explanation

$n(H) = 10$ {people who speak hindi}

$n(E) = 8$ {People who speak English}

$n(E\cup H) = 15$ (total)

$n (E \cap H) = 10 + 8 – 15 = 3$ (Speaking both)

People who speak hindi only $= 10 – 3 = 7$

Total ways of selecting $2$ people $= 15C_{2} = 105$

Favourable ways $= 7C_{1} \times 3C_{1} = 21$

Probability of occurrence $= \dfrac {21}{105} = \dfrac {1}{5}$

The value of

$^{50}C_4+\displaystyle\sum^{6}_{r=1}$

$^{56-r}C_3$ is?

0%
$^{56}C_4$
0%
$^{56}C_3$
0%
$^{55}C_3$
0%
$^{55}C_4$

Explanation

We need to find value of

$^{50}C_4+\sum_{r=1}^6\text{ } ^{56-r}C_3$

$=^{50}C_4+^{50}C_3+^{51}C_3+^{52}C_3+^{53}C_3+^{54}C_3+^{55}C_3$

$(\because ^nC_i+^nC_{i-1}=^{n+1}C_i)$

$=^{51}C_4+^{51}C_3+^{52}C_3+^{53}C_3+^{54}C_3+^{55}C_3$

$=^{52}C_4+^{52}C_3+^{53}C_3+^{54}C_3+^{55}C_3$

$=^{53}C_4+^{53}C_3+^{54}C_3+^{55}C_3$

$=^{54}C_4+^{54}C_3+^{55}C_3$

$=^{55}C_4+^{55}C_3$

$=^{56}C_4$

A telephone number

$d_1d_2d_3d_4d_5d_6d_7$ is called memorable if the prefix sequence

$d_1d_2d_3$ is exactly the same as either of the sequence

$d_4d_5d_6$ or

$d_5d_6d_7$ (or possibly both). If each

$d_1\epsilon\{x|0\leq x\leq 9, x\epsilon W\}$ , then number of distinct memorable telephone number is(are).

0%
$19810$
0%
$19,910$
0%
$19,990$
0%
$20,000$

$\sum _{ r=0 }^{ n-1 }{ \cfrac { { _{ }^{ n }{ C } }_{ r } }{ { _{ }^{ n }{ C } }_{ r }+{ _{ }^{ n }{ C } }_{ r+1 } } }$ is equal to

0%
$\cfrac { n }{ 2 }$
0%
$\cfrac { n+1 }{ 2 }$
0%
$\cfrac { n(n+1) }{ 2 }$
0%
$\cfrac { n(n-1) }{ 2(n+1) }$

Explanation

$\sum_{r=0}^{n-1}$

$\dfrac{^nC_r}{^nC_r+^nC_{r+1}}$

$\sum_{r=0}^{n-1}$

$\dfrac{^nC_r}{^{n+1}C_{r+1}}$ ----------(because

$^nC_r + ^nC_{r-1}=^{n+1}C_r$ )

$\sum_{r=0}^{n-1}\dfrac{^nC_r}{\dfrac{(n+1)}{(r+1)}^nC_r}$

$\sum_{r=0}^{n-1}\dfrac{r+1}{n+1}$

$\dfrac{n(n+1)}{2(n+1)}$

$\dfrac{n}{2}$

hence option

$'A'$ is correct.

A man

$x$ has

$7$ friends,

$4$ of them are ladies and

$3$ are men. His wife

$Y$ also has

$7$ friends,

$3$ of them are ladies and

$4$ are men. Assume

$X$ and

$Y$ have no common friends. Then, the total number of ways in which

$X$ and

$Y$ together can throw a party inviting

$3$ ladies and

$3$ men , so that

$3$ friends of each of

$X$ and

$Y$ are in this party, is

0%
$485$
0%
$468$
0%
$469$
0%
$484$

Explanation

Case 1: One of them invites all three the same gender.

ie,

$N=(^4C_3\ ^4C_3)+(^3C_3\ ^3C_3)=4^2+1^2=17$

Case 2: One of them invites 1 male and 2 females

$\Rightarrow$ The other invites 2 males and 1 female.

ie,

$N=(^4C_2\ ^3C_1\ ^4C_2\ ^3C_1)+(^4C_1\ ^3C_2\ ^3C_2\ ^4C_1)=18^2+12^2=468$

So, Total Possible ways

$=468+17=485$

ie, Option A is correct answer.

The sum

$^{20}C_0+^{20}C_1+^{20}C_2+..... +^{20}C_{10}$ is equal to

0%
$2^{20}+\dfrac{20 \,!}{(10\,!)^2}$
0%
$2^{19}+\dfrac{1}{2}.\dfrac{20\,!}{(10\,!)^2}$
0%
$2^{19}+^{20}C_{10}$
0%
none of these

Explanation

Let

$S={}^{ 20 }{ { C }_{ 0 } }+{}^{ 20 }{ { C }_{ 1 } }+{}^{ 20 }{ { C }_{ 2 } }+.......+{}^{ 20 }{ { C }_{ 10 } }$

$\Rightarrow 2S=\left( {}^{ 20 }{ { C }_{ 0 } }+{}^{ 20 }{ { C }_{ 0 } } \right) +\left( {}^{ 20 }{ { C }_{ 1 } }+{}^{ 20 }{ { C }_{ 1 } } \right) +\left( {}^{ 20 }{ { C }_{ 2 }+ }{}^{ 20 }{ { C }_{ 2 } } \right) +.......+\left( {}^{ 20 }{ { C }_{ 9 } }+{}^{ 20 }{ { C }_{ 9 } } \right) +\left( {}^{ 20 }{ { C }_{ 10 } }+{}^{ 20 }{ { C }_{ 10 } } \right)$

$\Rightarrow 2S-{}^{ 20 }{ { C }_{ 10 } }=\left( {}^{ 20 }{ { C }_{ 0 } }+{}^{ 20 }{ { C }_{ 20 } } \right) +\left( {}^{ 20 }{ { C }_{ 1 } }+{}^{ 20 }{ { C }_{ 19 } } \right) +\left( {}^{ 20 }{ { C }_{ 2 }+ }{}^{ 20 }{ { C }_{ 18 } } \right) +.......+\left( {}^{ 20 }{ { C }_{ 9 } }+{}^{ 20 }{ { C }_{ 11 } } \right) +{}^{ 20 }{ { C }_{ 10 } }$

(Using

$_{ }{}^{ n }{ { C }_{ r } }=_{ }{}^{ n }{ { C }_{ n-r } }$ )

$\Rightarrow 2S-{}^{ 20 }{ { C }_{ 10 } }={}^{ 20 }{ { C }_{ 0 } }+{}^{ 20 }{ { C }_{ 1 } }+{}^{ 20 }{ { C }_{ 2 }+ }.......+{}^{ 20 }{ { C }_{ 19 } }+{}^{ 20 }{ { C }_{ 20 } }$

$\Rightarrow 2S-{}^{ 20 }{ { C }_{ 10 } }={ 2 }{}^{ 20 }$ (Using

${}^{ n }{ { C }_{ 0 } }+{}^{ n }{ { C }_{ 1 } }+{}^{ n }{ { C }_{ 2 } }+.......+{}^{ n }{ { C }_{ n } }={ 2 }{}^{ n }$ )

$\Rightarrow S=\dfrac { { 2 }{}^{ 20 }+{}^{ 20 }{ { C }_{ 10 } } }{ 2 }$

$\Rightarrow S={ 2 }{}^{ 19 }+\dfrac { 1 }{ 2 } .\dfrac { 20! }{ { (10!) }{}^{ 2 } }$

Hence, Option B is correct.

The value of

$\begin{pmatrix} 30 \\ 0 \end{pmatrix}\begin{pmatrix} 30 \\ 10 \end{pmatrix}-\begin{pmatrix} 30 \\ 1 \end{pmatrix}\begin{pmatrix} 30 \\ 11 \end{pmatrix}+\begin{pmatrix} 30 \\ 2 \end{pmatrix}\begin{pmatrix} 30 \\ 12 \end{pmatrix}-...+\begin{pmatrix} 30 \\ 20 \end{pmatrix}\begin{pmatrix} 30 \\ 30 \end{pmatrix}$ is where

$\begin{pmatrix} n \\ r \end{pmatrix}={ _{ }^{ n }{ C } }_{ r }$

0%
$\begin{pmatrix} 30 \\ 15 \end{pmatrix}$
0%
$\begin{pmatrix} 60 \\ 15 \end{pmatrix}$
0%
$\begin{pmatrix} 60 \\ 30 \end{pmatrix}$
0%
$\begin{pmatrix} 30 \\ 10 \end{pmatrix}$

Explanation

$\displaystyle^{30}C_{0}\displaystyle^{30}C_{10}-\displaystyle^{30}C_{1}\displaystyle^{30}C_{11}+........+\displaystyle^{30}C_{20}\displaystyle^{30}C_{30}$

WKT

$\Rightarrow (x+1)^{30}=\displaystyle^{30}C_{0}+x\left(^{30}C_{1}\right)+x^2\left(^{30}C_{2}\right)+.........+x^{30}\left(^{30}C_{30}\right)$

$\Rightarrow (x-1)^{30}=x^{30}\left({^{30}C_{0}}\right)-x^{29}\left({^{30}C_{1}}\right)+......+^{30}C_{30}$

Multiply the above

$2$ equations

$\xi$ comparing

$x^{20}$ co-efficient.

$\Rightarrow (x+1)^{30}(x-1)^{30}=\left(x^2-1\right)^{30}$

$\Rightarrow T_{10+1}=^{30}C_{10}x^{2(10)}\times 1^{20}={^{30}C_{10}}x^{20}$

$\Rightarrow {^{30}C_{10}}={^{30}C_{0}}{^{30}C_{10}}-{^{30}C_{1}}{^{30}C_{11}}+.......+{^{30}C_{20}}{^{30}C_{30}}$

$\therefore$ Required

$={^{30}C_{10}}$

Hence, the answer is

${^{30}C_{10}}.$

Let

${ T }_{ n }$ denotes the number of triangles which can be formed by using the vertices of a regular polygon of n sides. If

${ T }_{ n+1 }\ -\ { T }_{ n }=21$ , then

$n$ is equal to

0%
$5$
0%
$7$
0%
$6$
0%
$4$

Explanation

Number of triangle that can be formed using polygon of

$n$ sided

$={}^{ n }{ C }_{3}$

According to question,

${}^{ n+1 }{ C }_{3}-{ }^{ n }{ C }_{3}=21$

$\dfrac { (n+1)(n)(n-1) }{ 6 } -\dfrac { n(n-1)(n-2) }{ 6 } =21$

$\Rightarrow \dfrac { 3(n)(n-1) }{ 6 } =21$

$\Rightarrow (n)(n-1)=42$

Solving this equation we get

$n=-6,7$

But

$n$ can't be negative

Hence,

$n=7$ .

$15! =2^{\alpha}\cdot 3^{\beta}\cdot 5^{\gamma}\cdot 7^{\delta}\cdot 11^{\theta}\cdot 13^{\Phi}$ , then the value of expression

$\alpha -\beta +\gamma -\delta +\theta -\Phi$ is

0%
$4$
0%
$6$
0%
$8$
0%
$10$

Explanation

$15!=1\times 2\times 3\times 4 \times 5\times 6\times 7\times 8\times 9\times 10\times 11\times 12\times 13\times 14\times 15$

$15!=2\times 3\times 2^2 \times 5\times 3\times 2 \times 7\times 2^3\times 3^2\times 2\times 5\times 11\times 3\times 2^2 \times 13\times 2\times 7 \times 3\times 5$

$15!=2^{11}\times 3^6\times 5^3\times 7^ 2\times 11\times 13$

$\implies \alpha =11$

$\implies \beta =6$

$\implies \gamma=3$

$\implies \delta =2$

$\implies \theta =1$

$\implies \Phi=1$

$\alpha-\beta+\gamma-\delta+\theta-\Phi=11-6+3-2+1-1=6$

$\left( 2\le r\le n \right)$ , then

${ _{ }^{ n }{ C } }_{ r }+2{ _{ }^{ n }{ C } }_{ r+1 }+{ _{ }^{ n }{ C } }_{ r+2 }$ is equal to

0%
$2.{ _{ }^{ n+2 }{ C } }_{ r+2 }\quad$
0%
${ _{ }^{ n+r }{ C } }_{ r+1 }$
0%
${ _{ }^{ n+2 }{ C } }_{ r+2 }$
0%
${ _{ }^{ n+1 }{ C } }_{ r }$

Explanation

To solve this, we'll use the relation

$^nC_{r}+^nC_{r+1}=^{n+1}C_{r+1}$ ...[1]

$^nC_{r}+2^nC_{r+1}+^nC_{r+2}$

$=(^nC_{r}+^nC_{r+1})+(^nC_{r+1}+^nC_{r+2})$

$= ^{n+1}C_{r+1}+^{n+1}C_{r+2}$ ...(Using [1])

$=^{n+2}C_{r+2}$ ...(Using [1])

From a collection of

$20$ consecutive natural numbers, four are selected such that they are not consecutive. The number of such selections is

0%
$284\times 17$
0%
$285\times 17$
0%
$284\times 16$
0%
$285\times 16$

Explanation

$1,2,3,4......20$

there are

$17$ way for four consecutive number

$\{(1,2,3,4),(2,3,4,5),(3,4,5,6),\dots \dots ,(17,18,19,20)\}$

numbers ways

$={ _{ }^{ 20 }{ C } }_{ 4 }-17=285\times 17-17=284\times 17$

The number of selection of

$n$ objects from

$2n$ objects of which

$n$ are identical and the rest are different is

0%
${ 2 }^{ n }$
0%
${ 2 }^{ n-1 }$
0%
${ 2 }^{ n }-1\quad$
0%
${ 2 }^{ n }+1\quad$

Explanation

Let

$S_1$ be the set of

$n$ identical objects and

$S_2$ be the set of the remaining

$n$ different objects.

Number of ways to select

$r$ objects from

$S_1=1$ since all objects in

$S_1$ are identical.

Number of ways to select

$r$ objects from

$S_2= ^nC_{r}$ .

Total number of ways to select

$n$ objects =

$\bullet$

$n$ objects from

$S_2$ and

$0$ object from

$S_1$

Number of ways

$= ^nC_{n}$

$\bullet$

$n-1$ objects from

$S_2$ and

$1$ object from

$S_1$

Number of ways

$= ^nC_{n-1}\times 1=^nC_{n-1}$

$\bullet$

$n-2$ objects from

$S_2$ and

$2$ object from

$S_1$

Number of ways

$= ^nC_{n-2}\times 1=^nC_{n-2}$

$\bullet$

$n-3$ objects from

$S_2$ and

$3$ object from

$S_1$

Number of ways

$= ^nC_{n-3}\times 1=^nC_{n-3}$

$.$

$\bullet$

$2$ objects from

$S_2$ and

$n-2$ object from

$S_1$

Number of ways

$= ^nC_{2}\times 1=^nC_{2}$

$\bullet$

$1$ objects from

$S_2$ and

$n-1$ object from

$S_1$

Number of ways

$= ^nC_{1}\times 1=^nC_{1}$

$\bullet$

$0$ objects from

$S_1$ and

$n$ object from

$S_1$

Number of ways

$= ^nC_{0}\times 1=^nC_{0}$

$\therefore$ Total number of ways

$= ^nC_{n}+^nC_{n-1}+^nC_{n-2}+^nC_{n-3}+....+^nC_{2}+^nC_{1}+^nC_{0}=2^n$

So, the answer is option (A).

The number of ways in which

$6$ rings can be worn on the four fingers of one hand is

0%
$4^{6}$
0%
$^{6}C_{4}$
0%
$6^{4}$
0%
None of these

Explanation

Each ring can be worn on

$4$ fingers, means each have

$4$ possibilities.

$\therefore$ Total ways

$=4\times4\times4\times4\times4\times4=4^{6}$

Hence,

$(A)$

Ramesh number of ways in which the letters of the word RAMESH can be placed in the squares of the given figure so that no row remains empty, is

0%
17280
0%
18720
0%
15840
0%
14400

Explanation

Case 1

$2.2.2$ from top to bottom

$(\ ^{6}C_{2}\times 2!)\times (\ ^{4}C_{2}\times 2!)\times (4\times 3)=4320$

Case 2

$2, 1, 3$ from top the bottom:

$(\ ^{6}C_{2}\times 2!)\times (\ ^{4}C_{1}\times 2)\times (4\times 3\times 2)=5760$

Case 3:

$1, 2, 3$ from top of bottom

$(\ ^{6}C_{1}\times 2)\times (\ ^{5}C_{2}\times 2!)\times (4\times 3\times 2)=5760$

Case 4:

$1,1,4$ from top the bottom

$(\ ^{6}C_{1}\times 2)\times (\ ^{5}C_{1}\times 2)\times (4!)$

So the correct answer is

$(B) 18, 720$

Let n be the number of ways in which the letters of the word "RESONANCE" can be arranged so the vowels appear at the even places and m be the number of ways in which "RESONANCE" can arrange so that letters R,S,O,A appears in the order same as the word RESONANCE, then answers the following questions.
The value of n is

0%
360
0%
720
0%
240
0%
840

Explanation

RESONANCE

VOWELS

$=E, O, A, E$

CONSTANTS

$=k, S, N, N, C$

$\bar{1}\ \bar{(2)}\ \bar{3}\ \bar{(4)}\ \bar{5}\ \bar{(6)}\ \bar{7}\ \bar{(8)}\ \bar{9}$

$\Rightarrow$ vowels can be arranged as

$=\dfrac{4!}{2!}$

$\Rightarrow$ consonants can be as

$=\dfrac{5!}{2!}$

Total No. of ways

$\dfrac{4!}{2!}\times \dfrac{5!}{2!}=\dfrac{24}{2}\times \dfrac{120}{2}=12\times 60 =720$

For

$n$ being natural number, if

$^{2n}C_r={}^{2n}C_{r+2}$ , find

$r$ .

0%
$n$
0%
$n-1$
0%
$n-2$
0%
$n-3$

Explanation

Given,

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

or,

$^{2n}C_{2n-r}=^{2n}C_{r+2}$ [ As

$^nC_r=^nc_{n-r}$ ]

or,

$2n-r=r+2$

or,

$2r=2n-2$

or,

$r=n-1$ .

How many different

$4$ -person committees can be chosen form the

$100$ members of the Senate ?

0%
$25$
0%
$400$
0%
$3,921,225$
0%
$94,109,400$

Explanation

Total number of members

$=100$ .
Members has to be selected

$=4$ .

$\therefore$ different committees

$=^{100}C_{4}=\large{\frac{100!}{4!\times96!}}$

$=3,921,225$ .

If the letterss of the word 'NAAGI' are arranged as in a dictionary then the rank of the given word is

0%
$23$
0%
$84$
0%
$49$
0%
$48$

Explanation

In a dictionary, letters in alphabetical order are

$A,A,G,I,N$

so the first word will be

$AAGIN$

Number of words beginning with

$A=$ Number of ways arranging

$A,G,I,N$

$=4!=24$

Number of words beginning with

$G=\dfrac{4!}{2!}=12$

Number of words beginning with

$I=\dfrac{4!}{2!}=12$

Next word will be

$NAAGI$

No. of words before

$NAAGI$

$=24+12+12=48$

So rank of the word

$NAAGI$ is

$49$ .

The number of ways in which we can select 5 letters of the word INTERNATIONAL is equal to

0%
$200$
0%
$220$
0%
$242$
0%
$256$

Explanation

We have thirteen letters.

$2l, 3N, 2T, 2A$ and

$(E, O, L,R)8$ (types)
We can select 5(five) letters in the following manner:

1. All different

${^8C}_5 = \dfrac{8.7.5}{1.2.3} = 56$ ways

2. 2 alike, 3 different

${^4C}_1 . {^7C}_3 = 435 = 140$ ways (we have 4 sets of alike letters)

3. 3 alike, 2 different

${^1C}_1 . {^7C}_2 = 1.21 = 21$ ways (we have only one set of 3 alike)

4. 3 alike and 2 alike

${^1C}_1 . {^3C}_1 = 1.3 = 3$ ways

5.Two sets of alike and one different

${^4C}_2 . {^6C}_1 = 6.6 = 36$ ways

$\therefore$ Total number of selection is

$56 + 140 + 21 + 3 + 36 = 256$ ways

$\Longrightarrow$ (d)

$^nC_r : ^nC_{r + 1} : ^nC_{r + 2} = 3 : 4: 5$ , then the value of

$2n + 3r$ is

0%
$238$
0%
$220$
0%
$203$
0%
$240$

Explanation

$\dfrac { { { n }_{ C } }_{ r } }{ { { n }_{ C } }_{ r+1 } } =\dfrac { 3 }{ 4 }$

$\dfrac { r+1 }{ n-r } =\dfrac { 3 }{ 4 }$

$\dfrac { { { n }_{ C } }_{ r+1 } }{ { { n }_{ C } }_{ r+2 } } =\dfrac { 4 }{ 5 }$

$\dfrac { r+2 }{ n-r-1 } =\dfrac { 4 }{ 5 }$

$On\quad solving\quad above\quad we\quad get\quad r=26,n=62$

$2n+3r=238$

$\dfrac { { { n }_{ C } }_{ r } }{ { { n }_{ C } }_{ r+1 } } =\dfrac { 3 }{ 4 }$

$\dfrac { r+1 }{ n-r } =\dfrac { 3 }{ 4 }$

$\dfrac { { { n }_{ C } }_{ r+1 } }{ { { n }_{ C } }_{ r+2 } } =\dfrac { 4 }{ 5 }$

$\dfrac { r+2 }{ n-r-1 } =\dfrac { 4 }{ 5 }$

$On\quad solving\quad above\quad we\quad get\quad r=26,n=62$

$2n+3r=238$

The total number of different combinations of one
or more letters which can be made from the letter of the word MISSISSIPPI is,

0%
$150$
0%
$148$
0%
$149$
0%
$146$

We wish to select

$6$ persons from

$8$ , but if the person A is chosen, then B must be chosen. In how many ways can selections be made?

0%
$20$
0%
$22$
0%
$24$
0%
$26$

Explanation

When A is not Taken then B may be Chosen (No of remaining Candidates

$=8-1=7$ )

No.of selections

$={}^7C_6=7$

When A is taken B must be chosen (No of remaining Candidates

$=6$ and number of remaining selections

$=4$ )

No. of selections

$={}^6C_4=15$

Total no. of ways

$=15+7=22$

Number of arrangements of the letter

$HOLLYWOOD$ in which all

$Os$ are separated.

0%
$360\ ^{ 7 }{ C }_{ 3 }$
0%
$15\ ^{ 7 }{ P }_{ 4 }$
0%
$15\ ^{ 7 }{ C }_{ 4 }$
0%
$30\ ^{ 7 }{ P }_{ 4 }$

$\dfrac { { C }_{ 1 } }{ { C }_{ 0 } } +2\dfrac { { C }_{ 2 } }{ { C }_{ 1 } } +3\dfrac { { C }_{ 3 } }{ { C }_{ 2 } } +....+15\dfrac { { C }_{ 15 } }{ { C }_{ 14 } }$ is equal to

0%
$100$
0%
$120$
0%
$-120$
0%
$None\ of\ these$

Explanation

$\begin{array}{l} \frac { { { C_{ 1 } } } }{ { { C_{ 0 } } } } +\frac { { 2{ C_{ 2 } } } }{ { { C_{ 1 } } } } +\frac { { B{ C_{ 3 } } } }{ { { C_{ 2 } } } } ........15\frac { { { C_{ 13 } } } }{ { { C_{ 14 } } } } \\ \frac { { { C_{ 1 } } } }{ { { C_{ 0 } } } } =4\, \\ \frac { { 2{ C_{ 2 } } } }{ { { C_{ 1 } } } } =n-1 \\ \frac { { 15{ C_{ 15 } } } }{ { { C_{ 14 } } } } =1 \\ n+\left( { n-1 } \right) ....1 \\ =\frac { { 15\left( { 15+1 } \right) } }{ 2 } \\ =120 \end{array}$

$\dfrac { ^{ n }{ P }_{ r-1 } }{ a } =\dfrac { ^{ n }{ P }_{ r } }{ b } =\dfrac { ^{ n }{ P }_{ r+1 } }{ c }$ , then which of the following holds good:

0%
$c^{2}=a(b+c)$
0%
$a^{2}=c(a+b)$
0%
$b^{2}=c(a-b)$
0%
$\dfrac {1}{a}+\dfrac {1}{b}+\dfrac {1}{c}=1$

Explanation

$\dfrac{^{n} {P}_{r-1}}{a} = \dfrac{^{n}{P}_{r}}{b} = \dfrac{^{n}{P}_{r+1}}{c}$

$\implies \dfrac{n!}{(n - r+ 1)!a} = \dfrac{n!}{(n-r)!b} = \dfrac{n!}{(n-r-1)!c}$

$\implies \dfrac{n!}{(n - r+ 1)!a} = \dfrac{n!}{(n-r)(n-r+1)!b} = \dfrac{n!}{(n-r-1)(n-r)(n-r+1)!c}$

$\implies \dfrac{1}{a} = \dfrac{1}{(n-r)b} = \dfrac{1}{(n-r-1)(n-r)c}$

$\implies n-r = \dfrac{a}{b}$ and

$n-r-1 = \dfrac{b}{c}$

$\implies n-r - 1 = \dfrac{a}{b} - 1$

$\implies \dfrac{b}{c} = \dfrac{a}{b} - 1$

$\implies \dfrac{b}{c} - \dfrac{a}{b} = - 1$

$\implies b^2 - ac = - bc$

$\implies b^2 = ac - bc$

$\therefore b^2 = c(a - b)$ (Ans)

How many committee of five persons with a chairperson can be selected from

$12$ persons.

0%
$924$
0%
$825$
0%
$736$
0%
$643$

Explanation

No of Persons

$=5+1$ (Chairperson)

$\therefore$ No of committees that can be selected

$={}^{12}C_6=924$

The number of permutations of the letters of the word

$HONOLULU$ taken

$4$ at a time is

0%
$354$
0%
$314$
0%
$210$
0%
$124$

Explanation

$H-1$

$N-1$

$O-2$

$L-2$

$U-2$

Number of permutation

$=\dfrac {n!}{r_1\times r_2\times....\times r_n} =\dfrac { 8! }{ { (2! })^{ 3 }4! } =210$

The English alphabet has 5 vowels and 21 consonants. How many words with two different vowels and 2 different consonants can be formed from the alphabet ?

0%
$50000$
0%
$50100$
0%
$50300$
0%
$50400$

Explanation

$2$ vowels can be chosen in

$5{c}_{2}$ ways.

$2$ consonants can be chosen in

$21{c}_{2}$ ways.

$4$ letter can be arranged in

$4$ ways.

$\therefore$ The no of words containing

$2$ vowels and

$2$ consonants

$=5{c}_{2}\times 21{c}_{2}\times 4!=10\times 210\times 24=50400$

Find the remainder when

$105!$ is divided by

$214.$

0%
$168$
0%
$108$
0%
$196$
0%
$172$

PQRS is a quadrilateral having

$3, 4, 5, 6$ points in PQ, QR, RS and SP respectively. The number of triangles with vertices on different sides is?

0%
$220$
0%
$270$
0%
$282$
0%
$342$

Explanation

The number of triangles with vertices on side,

$AB,BC,CD=3{C}_{1}\times 4C_1\times 5C_1$

$\therefore$ Total no. of triangles

$=3C_1\times 4C_1\times 5C_1+3C_1\times 4C_1\times 6C_1+3C_1\times 5C_1\times 6C_1+ 4C_1\times 5C_1\times 6C_1\\=3\times 4\times 5+3 \times 4\times 6+3 \times 5\times 6+4\times 5\times 6\\=342$

How many

$3$ -digit even numbers can be formed from the digits

$1, 2, 3, 4, 5, 6$ if the digits can be repeated?

0%
108
0%
98
0%
72
0%
112

Explanation

Only

$3$ numbers are possible at units place

$(2,4,6)$ as we need even numbers. But at

$10's$ place and

$100's$ place all

$6$ are possible.

No. of digit even no's

$=3\times 6\times 6=108$

1041092_1058820_ans_ef5ae83b4b1b430696b691084317186f.png

The value of

$\sum\limits_{r = 1}^{10} r .\frac{{{}^n{C_r}}}{{{}^n{C_{r - 1}}}}$ is equal to

0%
$5(2n-9)$
0%
$10n$
0%
$9(n-4)$
0%
$n-2$

Explanation

CORRECTION:

$10n-45 = 5(2n-9)$ Option A correct
1118371_1086652_ans_825f6be686f44f75b5b5749fdb7da7d3.jpg

Robert was asked to made a

$5$ digit number from the digits

$2$ and

$4$ such that first digit cannot be

$4$ . Find the number of such

$5$ digit numbers that can be formed .

0%
$10$
0%
$12$
0%
$16$
0%
$18$
0%
None of these

Explanation

Total

$5$ digit numbers

$=1\times 2\times 2\times 2\times 2$

$=2^{4}$

$=16$

$C$ is correct

The number of

$5$ digit telephone numbers having least one of their digits repeated is

0%
$90,000$
0%
$100,000$
0%
$30,240$
0%
$69,760$

Explanation

Consider the given question,

Numbers with repetition

$=10\times 10\times 10\times 10\times 10$

$=1,00,000$

Numbers without repetition

$=10\times 9\times 8\times 7\times 6$

$=30240$

Number with one digit is repeated

$=100000-30240$

$=69760$

Hence, the number of 5 digit telephone numbers having least one of their digits repeated is

$69760$

If A and B are the sums of odd and even terms respectively in the expansion of

$(x+a)^n$ ,then

$(x+a)^{2n} -(x-a)^{2n}$ is equal to:

0%
$4(A+B)$
0%
$4(A-B)$
0%
$AB$
0%
$4AB$

Explanation

$A$ =sum of odd terms
Let n is odd,let total

$T_{n+1}$ terms

$A=T_1+T_3+...T_n$

$A=^nC_0+^nC_2+....+^nC_{n-1}$

$A=\dfrac{2^n}{2}=2^{n-1}$
similarly

$B=T_2+T_4+...T_{n+1}$

$B=^nC_1+^nC_3+....+^nC_n$

$B=\dfrac{2^n}{2}=2^{n-1}$

$A+B=2^n$

$(x+a)^{2n} -(x-a)^2n \Rightarrow (x+a+x-a)^n (x+a-x+a)^n$

$(2x)^n (2a)^n \Rightarrow 2^{2n} x^n a^n=4^n a^n x^n$
Considering this term being

$4^n$

$AB=\dfrac{2^{2n}}{4}$

$4AB=2^{2n}=4^n$
Therefore,

$(x+a)^{2n}-(x-a)^{2n}=4AB$

Number of ways in which

$7$ green bottles and

$8$ blue bottles can be arranged in a row if exactly

$1$ pair of green bottles is side by side is (Assume all bottles to be a like except for the colour).

0%
$84$
0%
$360$
0%
$504$
0%
None of the above

Explanation

Consider the two green bottles as one entity. Let's call it

$GG$ .

We will also refer to the green bottle as:

$G$ , and to the blue bottle as:

$B$ .

So now we have

$14$ units to arrange ( 1 green pair bottle + the remaining 5 green bottles + the 8 blue bottles):

$GG, G, G, G, G, G, B, B, B, B, B, B, B, B$

Now place the 8 blue bottles in a row, such that between each bottle has a gap before and after it:

_B_B_B_B_B_B_B_B_

Now we need to take the six green units (the 1 pair and 5 bottles) and place them in six of the nine gaps.

Therefore, the solution is:

$^9C_6 = 84 ways$ .

There are

$10$ points in a plane of which no

$3$ points are collinear and

$4$ points are concylic. No. of different circles that can be drawn through atleast

$3$ points of these given points is

0%
$\left( ^{ 8 }{ C }_{ 3 }-^{ 6 }{ C }_{ 3 } \right) +1$
0%
$\left( ^{ 10 }{ C }_{ 3 }-^{ 4 }{ C }_{ 3 } \right) +1$
0%
$\left( ^{ 6 }{ C }_{ 3 }-^{ 4 }{ C }_{ 3 } \right) +1$
0%
$\left( ^{ 4 }{ C }_{ 3 }-^{ 2 }{ C }_{ 3 } \right) +1$

Explanation

We need at least

$3$ points to form a circle:

$^{ 10 }{ C }_{ 3 }$

Out of these

$4$ points are already concyclic so we must remove

$^{ 4 }{ C }_{ 3 }-1$ cases.

$^{ 10 }{ C }_{ 3 }-\left( ^{ 4 }{ C }_{ 3 }-1 \right)$

Answer is:

$\left( ^{ 10 }{ C }_{ 3 }-^{ 4 }{ C }_{ 3 } \right) +1$

The sides AB, BC, CA of a triangle ABC have

$3,4$ and

$5$ interior points respectively on them. The number of triangles that can be constructed using these points as vertices is-

0%
$205$
0%
$210$
0%
$315$
0%
$216$

Explanation

$\rightarrow$ Suppose we had

$12$ points, none of them collinear, number of possible triangles

$=^{ 12 }{ C }_{ 3 }$

$\rightarrow$ If out of these

$3,4$ and

$5$ points were chosen separately to lie on each side of a triangle ABC, then number of possible triangles

$=^{ 12 }{ C }_{ 3 }-^{ 3 }{ C }_{ 3 }-^{ 5 }{ C }_{ 3 }-^{ 4 }{ C }_{ 3 }$

$=220-1-10-4$

$=205$

CBSE Questions for Class 11 Engineering Maths Permutations And Combinations Quiz 7 - 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 7 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

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