MCQ Questions for Class 11 Engineering Maths Permutations And Combinations Quiz 6

The value of

$\dfrac {(n + 2)! - (n + 1)!}{n!}$ is:

0%
$(n + 2)!$
0%
$(n + 1)!$
0%
$(n + 2)^{2}$
0%
$(n + 1)^{2}$
0%
$n$

Explanation

The value of

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

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

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

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

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

Six points were chosen on a circle and every possible chord was drawn. Two chords, which do not have the common points are named separately. How many pairs of separate chords exist in the situation described above?

0%
$26$
0%
$28$
0%
$30$
0%
$34$

Explanation

We have to choose six points on circle and join them for every possible chord

So total number of ways is

$6!$

But Two chord, which do not have common points i.e different chord

Hence Two chord has

$4$ points So

Number of ways of different or separately chord is

$\dfrac{6!}{4!}=30$

If 100! =

$\displaystyle { 2 }^{ a }{ 3 }^{ b }{ 5 }^{ c }{ 7 }^{ d }...$ , then

0%
$a=97$
0%
$\displaystyle b=\frac { 1 }{ 2 } \left( a+1 \right)$
0%
$\displaystyle c=\frac { 1 }{ 2 } b$
0%
$\displaystyle d=\frac { 1 }{ 3 } b$

Explanation

i) no. of 2's in 100! are

$\left [ \dfrac{100}{2} \right ]+\left [ \dfrac{100}{22} \right ]+\left [ \dfrac{100}{23} \right ]+\left [ \dfrac{100}{24} \right ]+\left [ \dfrac{100}{25} \right ]+\left [ \dfrac{100}{26} \right ]$

$= 50+25+12+6+3+1$

$= 97$ ie

$\boxed{a = 97}$

ii) no of 3's 100! are

$\left [ \dfrac{100}{3} \right ]+\left [ \dfrac{100}{3^{2}} \right ]+\left [ \dfrac{100}{3^{3}} \right ]+\left [ \dfrac{100}{3^{4}} \right ]$

$= 33+11+3+1 = \boxed{48 = 6}$

iii) no of 5's in 100 ! are

$\left [ \dfrac{100}{5} \right ]+\left [ \dfrac{100}{5^{2}} \right ] = 20+4 = \boxed{24 = c}$

iv) no of 7's in 100 ! are

$\left [ \dfrac{100}{7} \right ]+\left [ \dfrac{100}{7^{2}} \right ] = 14+2 = \boxed{16 =d}$

$\boxed{c = \dfrac{b}{2}}$ &

$\boxed{d = \dfrac{b}{3}}$

1108205_518574_ans_647006bd59e945a4ad68cddbe95cdad1.jpg

In how many different orders can five boys stand on a line?

0%
40
0%
50
0%
80
0%
120

Explanation

Five boys can stand in a line in

$5! = 120$ ways.

Let

$\boxed { n }$ be defined as

$\frac{(n+2)!}{(n-1)!}$ , what is the value of

$\frac{\boxed{7}}{\boxed {3}}$ ?

0%
4.4
0%
8.4
0%
12.4
0%
16.4
0%
20.4

Explanation

Given, box

$n$ is equal to

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

We get box

$7$ is equal to

$9 \times 8 \times 7 =504$
We get box

$3$ is equal to

$5 \times 4 \times 3 = 60$
Therefore, If we divide those two, we get

$\dfrac{504}{60} = 8.4$ .

$^6P_r = 360$ and

$^6C_r =15$ ,then find

$r?$

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

Explanation

$\dfrac{6!}{(6-r)!}=360$

$\dfrac{6!}{(6-r)r!}=15$

$r!=\dfrac{360}{15}=24$

$r=4$

Find the coefficient of the middle term of the expansion

$\left (x-\dfrac{1}{2y}\right)^{10}$ :

0%
$-\dfrac{63}{8}$
0%
$24$
0%
$-\dfrac{33}{4}$
0%
$-33$

Explanation

We need to find middle term of

$\left (x-\dfrac {1}{2y}\right)^{10}$

Total number of terms are

$11$ .
So, middle term will be the

$6^{th}$ term.

$\therefore t_{5+1}= ^{10}C_5 (-1)^5 x^{10-5} (\dfrac {1}{2y})^5=-\dfrac{63x^5}{8y^5}$
Hence, option A is correct.

A company has

$5$ men and

$6$ women. What are the number of ways of selecting a group of eight persons?

0%
$165$
0%
$185$
0%
$205$
0%
$225$

Explanation

The number of ways of selecting a group of eight is

5 men and 3 women=

$^5C_5 \times ^6C_3$ =20

4 men and 4 women=

$^5C_4 \times ^6C_4$ =75

3 men and 5 women=

$^5C_3 \times ^6C_5$ =60

2 men and 6 women=

$^5C_2 \times ^6C_6$ =10

Thus the total possible cases is 20+75+60+10=165.

Option A is correct answer.

The value of

${ }^{10}C_1 +{ }^{10}C_2 + { }^{10}C_3 + ... + { }^{10}C_9$ is

0%
$2^{10}$
0%
$2^{11}$
0%
$2^{10}-2$
0%
$2^{10}-1$

Explanation

Consider

$(1+x)^{10}={ }^{10}C_0x^{10}+{ }^{10}C_1x^9+{ }^{10}C_2x^8+.........+{ }^{10}C_9x+{ }^{10}C_{10}$

Now substitute

$x=1$ both sides

$(1+1)^{10}={ }^{10}C_0+{ }^{10}C_1+{ }^{10}C_2+.........+{ }^{10}C_9+{ }^{10}C_{10}$

$\Rightarrow 2^{10}=1+{ }^{10}C_1+{ }^{10}C_2+.........+{ }^{10}C_9+1$

$\therefore { }^{10}C_1+{ }^{10}C_2+{ }^{10}C_3+......+{ }^{10}C_9=2{ }^{10}-2$

Note:

${ }^{n}C_0=1$ and

${ }^nC_n=1$

$^{40}C_{n+7} = ^{40}C_{4n-2},$ then all the values of n are given by

0%
$28$
0%
$3,6$
0%
$3,7$
0%
$6$

Explanation

Here there are two conditions

$n+7=4n-2$

$\Longrightarrow 3n=9$

$\Longrightarrow n=3$

and,

$n+7+4n-2=40$

$\Longrightarrow 5n+5=40$

$5n=35$

$n=7$

There are two possible value of

$n$ . They are

$3$ and

$7$

Ten different letters of alphabet are given, words with five letters are formed with these given letters. Then the number of words which have at least one letter repeated

0%
$69760$
0%
$30240$
0%
$99748$
0%
$42386$

Explanation

$10$ letters

$5$ to be selected

Total number of

$5$ letter words

$={ 10 }^{ 5 }$

Number of words without repetition

$={}^{ 10 }{ P }_{ 5 }$

Number of words without at least one letter repeated

$=100000-(10\times 9\times 8\times 7\times 6)\\ =69760$

Hence the correct answer is

$69760$

If,

$\dfrac{1}{^5C_r}+\dfrac{1}{^6C_r} = \dfrac{1}{^4C_r}$ , then the value of

$r$ equals to

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

Explanation

Since, there is a

$^4 C_r$ in the expression, then

$r \le 4$

$\dfrac{1}{\:^{5}C_{r}}+\dfrac{1}{\:^{6}C_{r}}=\dfrac{1}{\:^{4}C_{r}}$

$\dfrac{\:^{6}C_{r}+\:^{5}C_{r}}{\:^{5}C_{r}.\:^{6}C_{r}}=\dfrac{1}{\:^{4}C_{r}}$

$\dfrac{\:^{5}C_{r}(\dfrac{2(6)-r}{6-r})}{\:^{5}C_{r}.\:^{6}C_{r}}=\dfrac{1}{\:^{4}C_{r}}$

$\dfrac{12-r}{6-r}=\dfrac{\:^{6}C_{r}}{\:^{4}C_{r}}$

$\dfrac{12-r}{6-r}=\dfrac{\dfrac{6!}{(6-r)!}}{\dfrac{4!}{(4-r)!}}$

$\dfrac{12-r}{6-r}=\dfrac{5\times 6}{(6-r)(5-r)}$

$(12-r)(5-r)=30$

$(12-r)(5-r)=10\times 3$

$12-r=10$ or

$r=2$ .. similarly

$5-r=3$ or

$r=2$ .
Hence

$r=2$ is a solution.
Thus the answer is

$r=2$ .

When listing the integers from

$1$ to

$1000$ , how many times the digit

$5$ be written?

0%
$297$
0%
$243$
0%
$300$
0%
$273$

Explanation

Given range of integer

$(1$ to

$1000)$

From

$1$ to

$100$

The number of

$5$ in unit place

$=10$

From

$(1 to 1000)$ there will be

$100$

$5's$ in unit place.

Similarly in

$100$ place there will be

$10$

$'5's'$

From

$1$ to

$1000$ There wold be

$100$

$'5's'$ in tens place

Also in

$100's$ place There would be

$100$

$'5's'$

Total Number of

$5's$

$\Rightarrow 100+100+100\\ \Rightarrow 300$

Hence the correct answer is

$300$

$^nC_8 = ^nC_{27}$ , then what is the value of

$n?$

0%
$35$
0%
$22$
0%
$28$
0%
$41$

Explanation

Given,

$\Rightarrow ^{ n }{ C }_{ 8 }=^{ n }{ C }_{ 27 }$

We know ,

$\Rightarrow ^{ n }{ C }_{ r }=^{ n }{ C }_{ n-r }\\ \Rightarrow r+(n-r)=n\\ \Rightarrow 8+27=n\\ \Rightarrow n=35\\ \\$

Option

$:A$

The exponent of

$18$ in

$200!$ , is

0%
$24$
0%
$46$
0%
$47$
0%
$48$

Explanation

Exponent of $18$ in $200!$

$\Rightarrow 18=(3)^{ 2 }\times 2$

Exponent of $P$ in $n!$

$=\left[ \dfrac { n }{ p } \right] +\left[ \dfrac { n }{ p^{ 2 } } \right] +.............\left[ \dfrac { n }{ p^{ K } } \right] \\ =P^{ K }\le n$

Exponential of $3$ in $n!$

$=\left[ \dfrac { 200 }{ 3 } \right] +\left[ \dfrac { 200 }{ 9 } \right] +\left[ \dfrac { 200 }{ 27 } \right] +\left[ \dfrac { 200 }{ 81 } \right] \\ =66+22+4+2\\ =94$

Exponent of $(3)^{ 2 }=48$

Exponent of $2$ in $n!$

$=\left[ \dfrac { 200 }{ 2 } \right] +\left[ \dfrac { 200 }{ 4 } \right] +\left[ \dfrac { 200 }{ 8 } \right] +\left[ \dfrac { 200 }{ 16 } \right] >>48$

Least exponent $=48$

Exponential of $18$ is $48$

Hence the correct answer is $48$

There are

$8$ true/false questions in an examination.The number of ways in which this questions can be answered, is

0%
$256$
0%
$1024$
0%
$16$
0%
$64$

Explanation

$8$ True or False

Each questions has two choices

Two questions has

$2\times 2$ choices

Similarly

$8$ questions have

${ (2) }^{ 8 }$ choices

$\Rightarrow 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\\ \Rightarrow 4\times \times 4\times 4\times 4\\ \Rightarrow 16\times 16$

$\Rightarrow 256$ Choices

Hence the correct answer is

$256$

The total number of five digit numbers the sum of whose digits is odd is

0%
$9\times 10^4$
0%
$\dfrac{9\times 10^4}{2}$
0%
$10^5$
0%
none of these

Explanation

$0,1,2,3,4,5,6,7,8,9$

$5$ Digit number

If sum of first

$4$ digits is even then last digit should be odd and if it is odd then it should be even

$=5$ Ways

Total numbers

$=9\times 10\times 10\times 10\times 5\\ =4.5\times 10^{ 4 }\\ =\dfrac { 9 }{ 2 } \times 10^{ 4 }$

Hence the correct answer is

$\dfrac{ 9 }{ 2 }\times 10^{ 4 }$

There are

$8$ men and

$10$ women and you need to form a committee of

$5$ men and

$6$ women. In how many ways can the committee be formed?

0%
$10420$
0%
$11420$
0%
$11760$
0%
None of these

Explanation

Out of $8$ men and $10$ women . The committee has to select $5$ men and $6$ women.

$^8C_5 \times ^{10}C_6 =11760$

A box contains

$2$ white balls,

$3$ black balls and

$4$ red balls.In how many ways can three balls can be drawn from the box if atleast one black ball is to be included in the draw?

0%
$32$
0%
$48$
0%
$64$
0%
$96$

Explanation

$2$ White balls

$3$ Black balls

$4$ Red balls

Three balls are drawn

Total possible combination

$\Rightarrow ^{ 9 }{ P }_{ 3 }$

Combination without black

$\Rightarrow ^{ 2 }{ C_{ 2 } }\times ^{ 4 }{ C }_{ 1 }+^{ 2 }{ C }_{ 1 }\times ^{ 4 }{ C }_{ 2 }+^{ 4 }{ C }_{ 3 }$

The combination with atleast one black

$\Rightarrow \dfrac { 9\times 8\times 7 }{ 3\times 2\times 1 } -\left[ 1\times 4+2\times \dfrac { 12 }{ 2 } +4 \right] \\ \Rightarrow 12(7)-\left[ 20 \right] \\ \Rightarrow 84-20=64$

$64$ Combination include atleast one black

Hence the correct answer is $64$

The total number of five digit numbers the sum of whose digits is even is

0%
$4.5\times {10}^4$
0%
$9\times{10}^4$
0%
${10}^5$
0%
none of these

Explanation

$0,1,2,3,4,5,6,7,8,9$

Five digit number

Last digit can be filled in

$5$ ways

That is if the first four digits add up to odd sum then last digit will be odd

Else it would be even

Total number of such digits

$=9\times 10\times 10\times 10\times 5\\ =45\times 10^{ 3 }$

$=4.5\times 10^{ 4 }$

Hence, the correct answer is

$4.5\times 10^{ 4 }$ .

How many quadrilaterals can be formed by joining the vertices of an octagon?

0%
$65$
0%
$60$
0%
$70$
0%
$64$

Explanation

Octagon Number of sides

$= n=8$

For quadrilatral there are only four sides

There should be a selection of four ponts and arranging of quadrilatral

$\Rightarrow ^{ 8 }{ C }_{ 4 }=\cfrac { 8! }{ 4!\times 4! } \\ \Rightarrow \cfrac { 8\times 7\times 6\times 5 }{ 4\times 3\times 2 } \\ \Rightarrow 70\quad quadrilatrals$

Option

$: C$

The number of values of

$r$ satisfying the equation

$\:^{69}C_{3r-1}-\:^{69}C_{r^2}=\:^{69}C_{r^2-1}-\:^{69}C_{3r}$ is:

0%
$1$
0%
$2$
0%
$4$
0%
$7$

Explanation

Solution:

${}^{69}C_{3r-1}-{}^{69}C_{r^2}={}^{69}C_{r^2-1}-{}^{69}C_{3r}$

or,

${}^{69}C_{3r-1}+{}^{69}C_{3r}={}^{69}C_{r^2}+{}^{69}C_{r^2-1}$

or,

${}^{70}C_{3r}={}^{70}C_{r^2}$

Here,

$3r=r^2$ or

$3r+r^2=70$

$(\because {}^n C_r={}^n C s$ then

$r=s$ or

$r+s=n)$

or,

$3r=r^2$

or,

$r(r-3)=0$

or,

$r=0$ which is not possible.

$r=3$

$r^2+3r-70=0$

or,

$(r-7)(r+10)=0$

or,

$r=7$ or

$r=-10$ which is not possible.

The value of

$r$ satisfying the equation

$=3,7$

Number of values of

$r$ satisfying the equation

$=2$

Hence, B is the correct option.

An event manager has ten patterns of chairs and eight patterns of tables. In how many ways can he make a pair of table and chair?

0%
$100$
0%
$80$
0%
$110$
0%
$64$

Explanation

There are

$10$ patterns of chairs and

$8$ patterns of tables.

A pair of table and chair is required .

Therefore

$=^{10}C_1 \times 8C_1=80$

Find the number of subsets of the set

${1,2,3,4,5,6,7,8,9,10,11}$ having

$4$ elements

0%
$330$
0%
$320$
0%
$310$
0%
$300$

Explanation

Number of digits

$=11$

$1,2,3,4,5,6,7,8,9,10,11$

$4$ Elements are to be chosen

Number of ways

$^{ 11 }{ C }_{ 4 }$

$=\cfrac { 11! }{ 4!\times 7! }$

$=\cfrac { 11\times 10\times 9\times 8 }{ 4\times 3\times 2 }$

$=330$

Number of subsets

$=330$

When two coins are tossed and a cubical dice is rolled, then the total outcomes for the compound event is

0%
$48$
0%
$52$
0%
$36$
0%
$24$

Explanation

Outcomes when

$2$ coin tossed

$=2^2=4$

Outcomes when dice is rolled

$=6$

According to basic multiplication principle the total outcomes of

$n$ independent events is calculated by multiplying outcomes of each and every individual outcomes.

$\Rightarrow$ Total outcome

$=4\times6=24.$

Hence, the answer is

$24.$

$^nC_{10} = ^nC_{12}$ , then find n?

0%
$120$
0%
$22$
0%
$12$
0%
$2$

Explanation

$^{ n }{ C }_{ 10 }=^{ n }{ C }_{ 12 }$

We know

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

$\Rightarrow r+(n-r)=n\\ =10+12=n\\ \Rightarrow n=22$

There fore the required value of

$n=22$

Simplify:

${ _{ }^{ 34 }{ C } }_{ 5 }+\sum _{ r=0 }^{ 4 }{ { _{ }^{ (38-r) }{ C } }_{ 4 } }$ .

0%
${ }^{38}C_4$
0%
${ }^{39}C_4$
0%
${ }^{38}C_5$
0%
${ }^{39}C_5$

Explanation

$\displaystyle { }^{34}C_5 + \sum_{r = 0}^{4}$

${ }^{38 - r} C_4$

$= { }^{34}C_5 + { }^{38}C_4 + { }^{37}C_4 + { }^{36}C_4 + { }^{35}C_4 + { }^{34}C_4$

$= ({ }^{34}C_5 + { }^{34}C_4) + { }^{35}C_4 + { }^{36}C_4 + { }^{37}C_4 + { }^{38}C_4$

$= { }^{35}C_5 + { }^{35}C_4 + { }^{36}C_4 + { }^{37}C_4 + { }^{38}C_4$ (Using

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

$= { }^{36}C_5 + { }^{36}C_4 + { }^{37}C_4 + { }^{38}C_4$

$= { }^{37}C_5 + { }^{37}C_4 + { }^{38}C_4$

$= { }^{38}C_5 + { }^{38}C_4$

$= { }^{39}C_5$

$C\left( 28,2r \right) =C\left( 28,2r-4 \right)$ , then what is

$r$ equal to?

0%
$7$
0%
$8$
0%
$12$
0%
$16$

Explanation

We know that,

${}^{n}{C}_{k}={}^{n}{C}_{n-k}$

Here

$n=28, k=2r$

$\Rightarrow n-k=2r-4$ .....

$[\because {}^{28}C_{2r}={}^{28}{C}_{2r-4}]$

$\Rightarrow 28-2r=2r-4$

$\Rightarrow 4r=32$

$\Rightarrow r=8$

Hence,

$r=8$

Out of

$7$ consonants and

$4$ vowels, words are formed each having

$3$ consonants and

$2$ vowels. The number of such words that can be formed is

0%
$210$
0%
$25200$
0%
$2520$
0%
$302400$

Explanation

Total no. of consonants is

$7$ and of vowels is

$4$

Out of

$7$ consonants,

$3$ are chosen to form a word. So, this can be done in

${ }^{7}C_{3}$ ways

Out of

$4$ vowels,

$2$ are chosen to form a word. So, that can be done in

${ }^{4}C_{2}$ ways

So, the no. of select

$3$ consonants and

$2$ vowels is

$^{7}C_{3} \times ^{4}C_{2}$

Now, we can arrange the word containing

$5$ letters in

$5!$ ways.

Thus, the total no. of words formed each having

$3$ consonants and

$2$ vowels is

$^{7}C_{3} \times ^{4}C_{2}\times 5! = 25200$

Hence, the answer is

$25200$ .

${ _{ }^{ n }{ C } }_{ 8 }={ _{ }^{ n }{ C } }_{ 5 }$ , then the value of

$n$ is

0%
$2$
0%
$3$
0%
$1$
0%
$13$

Explanation

Given,

${ _{ }^{ n }{ C } }_{ 8 }={ _{ }^{ n }{ C } }_{ 5 }$ .

We have the formula for combination,

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

$\implies$

$n-r=5$ , if

$r=8$ OR

$n-r=8$ , if

$r=5$

$\implies$

$n-8=5$ OR

$n-5=8$

From both the equations we get

$n=13$ .

Hence, option

$D$ is correct.

Out of

$15$ points in a plane, n points are in the same straight line,

$445$ triangles can be formed by joining these points. What is the value of n?

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

Explanation

Number of triangles that can be formed is equal to the number of ways to select 3 non-collinear points.

Number of ways to select 3 points from 15 points

$=^{15}C_3$

Let n points be collinear.

Number of ways to select 3 points out of the n collinear points

$=^{n}C_3$

So, Number of ways to select 3 non-collinear points = Number of ways to select 3 points using all the points - Number of ways to select 3 points using the collinear points

So, Number of ways to select 3 non-collinear points

$=^{15}C_3-^{n}C_3$

So, Number of triangles that can be formed

$=^{15}C_3-^{n}C_3$

$\therefore 445=^{15}C_3-^{n}C_3$

$\therefore 445=455-^{n}C_3$

$\therefore ^{n}C_3=10$

$\therefore \dfrac{n!}{3!(n-3)!}=10$

$\therefore \dfrac{n(n-1)(n-2)}{6}=10$

$\therefore n(n-1)(n-2)=60$

Solving this equation we get

$n=5$

The answer is option (C)

Given that C(n, r) : C ( n, r + 1)=1 : 2 and C (n, r + 1) : C (n, r + 2) = 2 : 3.

Find

$n.$

0%
$11$
0%
$12$
0%
$13$
0%
$14$

Explanation

Given :

$\cfrac { _{ r }^{ n }{ C } }{ _{ r+1 }^{ n }{ C } } =\cfrac { 1 }{ 2 }$

$\Rightarrow 2\cfrac { n! }{ r!(n-r)! } = \cfrac{n!}{(r+1)!(n-r-1)!}$

$\Rightarrow 3r+2=n$

Also,

$\cfrac { { _{ r+1 }^{ n }{ C } } }{ _{ r+2 }^{ n }{ C } } =\cfrac { 2 }{ 3 }$

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

$\Rightarrow 3r+6=2n-2r-2$

$\Rightarrow 5r+8=2n$

Substituting

$n$ , we get

$8+5r=2(3r+2)$

$8+5r=6r+4$

$\Rightarrow r=4$

Solving for

$n$ , we get

$n=3(4)+2=12+2$

$n=14$

Hence, D is correct.

If different words are formed with all the letters of the word 'AGAIN' and are arranged alphabetically among themselves as in a dictionary, the word at the 50th place will be

0%
NAAGI
0%
NAAIG
0%
IAAGN
0%
IAANG

Explanation

We have the word

$AGAIN$

The words start from

$AAGIN=4!=24$

The words start from

$GAAIN=\cfrac{4!}{2!}=12$

The words start from

$IAAGN=\cfrac{4!}{2!}=12$

Number of total words

$=48$

$\therefore 49th$ word is

$NAAGI$ and

$50th$ word is

$NAAIG.$

Hence, B is the correct option.

The value of

$(^{21}C_1-^{10}C_1)+(^{21}C_2-^{10}C_2)+(^{21}C_3-^{10}C_3)+(^{21}C_4+^{10}C_4)+......+(^{21}C_{10}-^{10}C_{10})$ is:

0%
$2^{21}-2^{11}$
0%
$2^{21}-2^{10}$
0%
$2^{20}-2^9$
0%
$2^{20}-2^{10}$

Explanation

$(^{21}C_1-^{10}C_1)+(^{21}C_2-^{10}C_2)+(^{21}C_3-^{10}C_3)+(^{21}C_4+^{10}C_4)+......+(^{21}C_{10}-^{10}C_{10})$

$= ^{21}C_1+^{21}C_2+^{21}C_3+ ..... + ^{21}C_{10} - [2^{10}-1]$

$=\cfrac 12 [2 ^{21}C_1 + 2 ^{21}C_2+ 2 ^{21}C_3 + ..... + 2 ^{21}C_{10}] - [2^{10}-1]$

$=\cfrac 12 [^{21}C_1 + ^{21}C_2+ .... + ^{21}C_{10} + ^{21}C_{11} + ..... + ^{21}C_{19} + ^{21}C_{20}] - [2^{10}-1]$

$=\cfrac 12 [2^{21} - 1 -1 ] - [2^{10}-1]$

$=2^{20} - 1 - 2^{10} +1$

$=2^{20}-2^{10}$

The number of six-digit numbers which have sum of their digits as an odd integer, is

0%
$45000$
0%
$450000$
0%
$97000$
0%
$970000$

Explanation

The first six digit no.

$=100000$
The last six digit no.

$=999999$
No. of six digit no.

$=999999-100000+1$

$=900000$

In every two consecutive numbers, one of them will have sum of their digits as even integer.

For example,

$100000$ have sum of their digits equal to

$1$

Then

$100001$ will have sum of their digits equal to

$2$ and so on...

So, every alternate number have sum of their digits as even number and similarly the other alternate pair of numbers have their sum of digits as odd integer.

Thus, the number of six-digit numbers which have sum of their digits as odd integer and the number of six-digit numbers which have sum of their digits as even integer are equal.

And we know that, sum of digits of the number is either odd or even.

Thus, the number of six-digit number having sum of their digits as odd integer is half of the total number we have

$\therefore$ Out of

$900000$ , half of them will have sum as odd.
Therefore

$450000$ six digit no. will have sum of their digits as an odd integer.

$^nC_{r-1}=10, ^nC_r=45$ and

$^nC_{r+1}=120$ , then

$r$ equals to

0%
$1$
0%
$2$
0%
$3$
0%
$4$

Explanation

Using

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

$\Rightarrow \dfrac{{}^nC_r}{{}^nC_{r-1}}=\dfrac{45}{10}$ and

$\dfrac{{}^nC_{r+1}}{{}^{n}C_{r}}=\dfrac {120}{45}$

$\Rightarrow \dfrac{n-r+1}{r}=\dfrac{9}{2}$
and

$\dfrac{n-r}{r+1}=\dfrac{8}{3}$

$\Rightarrow \dfrac{8}{3}(r+1)+1=\dfrac{9}{2}r$

$\Rightarrow 16 r+16+6=27 r$

$\Rightarrow 11r=22$

$\therefore r=2$

$^nC_{12} = ^nC_8$ then is equal to

0%
$12$
0%
$26$
0%
$6$
0%
$20$

Explanation

$^{ n }{ C }_{ 12 }=^{ n }{ C }_{ 8 }$

$\cfrac { n! }{ 12!(n-12)! } =\cfrac { n! }{ 8!(n-8)! }$

$\cfrac { 1 }{ 12\times 11\times 10\times 9\times 8!\left( n-12 \right) ! } =\cfrac { 1 }{ 8!(n-8)\times (n-9)(n-10)(n-11) (n-12)! }$

$\Longrightarrow \left( n-8 \right) \left( n-9 \right) \left( n-10 \right) \left( n-11 \right) =12\times 11 \times 10 \times 9$

$\therefore n-8=12$

$n=8+12$

$n=20$

The value of

$\left( \dfrac { ^{ 50 }{ C }_{ 0 } }{ 1 } +\dfrac {^{ 50 }{ C }_{ 2 } }{ 3 } +...+\dfrac {^{ 50 }{ C }_{ 50 } }{ 51 } \right)$ is

0%
$\dfrac { { 2 }^{ 50 } }{ 51 }$
0%
$\dfrac { { 2 }^{ 50 }-1 }{ 51 }$
0%
$\dfrac { { 2 }^{ 50 }-1 }{ 50 }$
0%
None of these

Explanation

$\left( \dfrac { _{ }^{ 50 }{ { C }_{ 0 } } }{ 1 } +\dfrac { _{ }^{ 50 }{ { C }_{ 2 } } }{ 3 } +...+\dfrac { _{ }^{ 50 }{ { C }_{ 50 } } }{ 51 } \right)$

$=\left( \dfrac { 1 }{ 1 } +\dfrac { 50\times 49 }{ 3\times 2! } +...+\dfrac { 1 }{ 51 } \right)$

$=\dfrac { 1 }{ 51 } \left( \dfrac { 51 }{ 1! } +\dfrac { 51\times 50\times 49 }{ 3! } +...+1 \right)$

$=\dfrac { 1 }{ 51 } \left( _{ }^{ 51 }{ { C }_{ 1 } }+_{ }^{ 51 }{ { C }_{ 3 } }+...+_{ }^{ 51 }{ { C }_{ 51 } } \right)$

$=\dfrac { 1 }{ 51 } { 2 }^{ 51-1 }=\dfrac { { 2 }^{ 50 } }{ 51 }$

The number of positive integers satisfying the inequality

${ _{ }^{ n+1 }{ C } }_{ n-2 }-{ _{ }^{ n+1 }{ C } }_{ n-1 }\le 50$ is

0%
$9$
0%
$8$
0%
$7$
0%
$6$

Explanation

${ _{ }^{ n+1 }{ C } }_{ n-2 }-{ _{ }^{ n+1 }{ C } }_{ n-1 }\le 50$

$\Rightarrow \cfrac { (n-1)! }{ 3!(n-2)! } -\cfrac { (n+1)! }{ 2!(n-2)! } \le 50$

$\Rightarrow (n+1)!\left( \cfrac { n-1-3 }{ (n-1)! } \right) \le 300\quad$

$\Rightarrow (n+1)n(n-4)\le 300$

By trail and error we get

$n\leq8$ to satisfy the above inequality.

Find the term independent of

$x$ in

${ \left( \cfrac { 3 }{ 2 } { x }^{ 2 }-\cfrac { 1 }{ 3x } \right) }^{ 9 }$ .

0%
$\dfrac {6}{15}$
0%
$\dfrac {7}{18}$
0%
$\dfrac {7}{8}$
0%
$\dfrac {4}{9}$

Explanation

We know that the general term in expansion of

$(a+b)^n$ is given by

$^nC_{r}a^{n-r}b^{r}$ where r ranges from

$0$ to

$9$ .

Here

$a= \cfrac{3}{2}x^2, \ b= -\cfrac{1}{3x} \ \And \ n=9$

For any term to be independent of

$x$ , coefficient of

$x$ in

$a^{n+1-r}b^{r-1}$ should be zero.

$\Rightarrow \ (x^2)^{9-r}(x^{-1})^{r}=x^0$

$\Rightarrow 18-2r-r=0$

$\Rightarrow r=6$

Therefore, the term independent of

$x$ is

$^9C_6 \left (\cfrac{3}{2}\right)^3 \left (-\cfrac{3}{3}\right)^6=\cfrac{7}{18}$ .

Write down and simplify:
The

$5^{th}$ term of

${ \left( \cfrac { { x }^{ \frac { 3 }{ 2 } } }{ { a }^{ \frac { 1 }{ 2 } } } -\cfrac { { y }^{ \frac { 5 }{ 2 } } }{ { b }^{ \frac { 3 }{ 2 } } } \right) }^{ 8 }$ .

0%

$\dfrac {60x^6y^{5}}{a^2b^6}$
0%
$\dfrac {70x^6y^{10}}{a^3b^5}$
0%
$\dfrac {70x^5y^{5}}{a^2b^6}$
0%
$\dfrac {70x^6y^{10}}{a^2b^6}$

Explanation

The

$r^{th}$ term in expansion of

$(p+q)^n$ is given by

$^nC_{r-1}p^{n+1-r}q^{r-1}$

$\therefore$ in the given expression

$p=\cfrac{x^\frac{3}{2}}{a^\frac{1}{2}}, \ q=-\cfrac{y^\frac{5}{2}}{b^\frac{3}{2}}, \ r=5 \ \And \ n=8$

$\therefore$ the

$5^{th}$ term will be

$^8C_4 { \left( \cfrac{x^\frac{3}{2}}{a^\frac{1}{2}}\right)^4 {\left( - \cfrac{y^\frac{5}{2}}{b^\frac{3}{2}} \right)}^4} = \cfrac{70x^6y^{10}}{a^2b^6}$

The number of positive integer satisfying the inequality

$^{n + 1}C_{n} - {}^{n + 1}C_{n - 1} \leq 100$ is

0%
$9$
0%
$8$
0%
$5$
0%
None of these

Explanation

$^{n + 1}C_{n } - {}^{n + 1}C_{n - 1} \leq 100$

$\Rightarrow {}^{n + 1}C_{3} - {}^{n + 1}C_{2} \leq 100$

$\Rightarrow \dfrac {(n + 1)n(n - 1)}{6} - \dfrac {(n + 1)n}{2} \leq 100$

$\Rightarrow (n + 1)n(n - 1) - 3n(n + 1)\leq 600$

$\Rightarrow (n + 1) n(n - 4)\leq 600$
The values of

$n$ satisfying this inequality are

$2, 3, 4, 5, 6, 7, 8, 9$ .

The number of triangles that can be formed by choosing the vertices from a set of

$12$ points of which

$7$ points lie on a line is

0%
$185$
0%
$175$
0%
$115$
0%
$105$

Explanation

We can choose any three points out of 12 points and then subtract the no. of ways choosing three points out of 7 points which are in straight line.

$\therefore$ Total no. of triangles

$= ^{12}C_3-^7C_3=185$

Find the

$4^{th}$ term of

${ \left( 9x-\cfrac { 1 }{ 3\sqrt { x } } \right) }^{ 18 }$ .

0%
$16500$
0%
$18564$
0%
$16540$
0%
$32600$

Explanation

We know that the

$r^{th}$ term in expansion of

$(a+b)^n$ is given by

$^nC_{r-1}a^{n+1-r}b^{r-1}$

Here

$a=9x, \ n=18 \ \And \ b=-\cfrac{1}{3\sqrt{x}}$

$\therefore$ the fourth term in expansion of

$\left (9x-\cfrac{1}{3\sqrt{x}}\right)^{18}$ is

$^{18}C_{12} {x}^{6}\left (-\cfrac{1}{3\sqrt{x}}\right)^{12} =18564$

$^{n}C_{r - 1} = 36$ and

$^{n}C_{r} = 84$ , then

0%
$13r - n - 3 = 0$
0%
$10r - 3n - 30 = 0$
0%
$10r + 3n - 3 = 0$
0%
$10r - 3n + 3 = 0$
0%
$10r - 3n - 3 = 0$

Explanation

Given,

$^{n}C_{r - 1} = 36$
and

$^{n}C_{r} = 84$
Now,

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

$\Rightarrow \dfrac {r}{n - r + 1} = \dfrac {36}{84} = \dfrac {3}{7}$

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

$\Rightarrow 10r = 3n + 3$

$\Rightarrow 10r - 3n - 3 = 0$ .

The arithmetic mean of

${ _{ }^{ n }{ C } }_{ 0 },{ _{ }^{ n }{ C } }_{ 1 },....{ _{ }^{ n }{ C } }_{ n }\quad$ , is

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

Explanation

Clearly,
Required mean

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

$\quad { _{ }^{ n }{ C } }_{ 2 }+{ _{ }^{ n }{ C } }_{ 3 }={ _{ }^{ 6 }{ C } }_{ 3 }$ and

${ _{ }^{ n }{ C } }_{ x }={ _{ }^{ n }{ C } }_{ 3 },x\neq 3$ then the value of

$x$ is

0%
$5$
0%
$4$
0%
$2$
0%
$6$
0%
$1$

Explanation

Given,

${ _{ }^{ n }{ C } }_{ 2 }+{ _{ }^{ n }{ C } }_{ 3 }={ _{ }^{ n }{ C } }_{ 3 }$

$\Rightarrow n=5\left[ \because { _{ }^{ n }{ C } }_{ r }+{ _{ }^{ n }{ C } }_{ r-1 }={ _{ }^{ n+1 }{ C } }_{ r } \right]$
and

$\quad { _{ }^{ n }{ C } }_{ x }={ _{ }^{ n }{ C } }_{ 3 }\Rightarrow { _{ }^{ 5 }{ C } }_{ x }={ _{ }^{ 5 }{ C } }_{ 3 }$

$\quad \Rightarrow x=5-3=2$

$\displaystyle \sum _{ k=0 }^{ 18 }{ \cfrac { k }{ { _{ }^{ 18 }{ C } }_{ k } } } =a\sum _{ k=0 }^{ 18 }{ \cfrac { 1 }{ { _{ }^{ 18 }{ C } }_{ k } } }$ , then the value of

$a$ is

0%
$3$
0%
$9$
0%
$6$
0%
$18$
0%
$36$

Explanation

Given

$\displaystyle \sum _{ k=0 }^{ 18 }{ \cfrac { k }{ { _{ }^{ 18 }{ C } }_{ k } } } =a\sum _{ k=0 }^{ 18 }{ \cfrac { 1 }{ { _{ }^{ 18 }{ C } }_{ k } } }$

$LHS= \displaystyle\sum _{ k=0 }^{ 18 }{ \cfrac { k }{ { _{ }^{ 18 }{ C } }_{ k } } } =\left[ 0+\cfrac { 1 }{ { _{ }^{ 18 }{ C } }_{ 1 } } +\cfrac { 1 }{ { _{ }^{ 18 }{ C } }_{ 2 } } +...+\cfrac { 1 }{ { _{ }^{ 18 }{ C } }_{ 18 } } \right] ...(i)\quad$

$RHS=a\displaystyle \sum _{ k=0 }^{ 18 }{ \cfrac { k }{ { _{ }^{ 18 }{ C } }_{ k } } } =a\left[ 1+\cfrac { 1 }{ { _{ }^{ 18 }{ C } }_{ 1 } } +\cfrac { 1 }{ { _{ }^{ 18 }{ C } }_{ 2 } } +...+\cfrac { 1 }{ { _{ }^{ 18 }{ C } }_{ 18 } } \right] ...(ii)\quad$
From Eq(i) and (ii)

$18\left( 1+\cfrac { 1 }{ { _{ }^{ 18 }{ C } }_{ 1 } } +\cfrac { 1 }{ { _{ }^{ 18 }{ C } }_{ 2 } } +...+\cfrac { 1 }{ { _{ }^{ 18 }{ C } }_{ 18 } } \right) +\cfrac { 9 }{ { _{ }^{ 18 }{ C } }_{ 9 } }$

$=2a\left( 1+\cfrac { 1 }{ { _{ }^{ 18 }{ C } }_{ 1 } } +\cfrac { 1 }{ { _{ }^{ 18 }{ C } }_{ 2 } } +...+\cfrac { 1 }{ { _{ }^{ 18 }{ C } }_{ 18 } } \right) +\cfrac { a }{ { _{ }^{ 18 }{ C } }_{ 9 } }$

$\therefore$

$2a=18$

$\Rightarrow$

$a=9$

There are

$10$ persons including

$3$ ladies. A committee of 4 persons including atleast one lady is to be formed. The number of ways of forming such a committee is :

0%
$160$
0%
$170$
0%
$180$
0%
$175$
0%
$155$

Explanation

A committee pf

$4$ persons can be formed in following ways:
(i)

$3$ persons and one ladies

$=^7C_3 \times ^3C_1 = 35 \times 3 = 105$
(ii)

$2$ persons and two ladies

$= ^7C_2 \times ^3C_2 = 21 \times 3 = 63$
(iii)

$1$ persons and three ladies

$= ^7C_1 \times ^3C_3 = 7 \times 1 = 7$
Therefore, total number of ways

$= 105 + 63 + 7 = 175$

If PQRS is a convex quadrilateral with 3, 4, 5 and 6 points marked on side PQ, QR, RS and PS respectively. Then, the number of triangles with vertices on different sides is

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

Explanation

Total number of triangles

$=$ Number of triangles with vertices on sides

$(\text{PQ, QR, RS + QR, RS, PS + RS, PS, PQ + PS, PQ, QR})$

$= {}^3C_1 \times {}^4C_1 \times {}^5C_1 + {}^4C_1 \times {}^5C_1 \times {}^6C_1 + {}^5C_1 \times {}^6C_1 \times {}^3 C_1 + {}^6C_1 \times {}^3C_1 \times {}^4C_1$

$= 60 + 120 + 90 + 72 = 342$

CBSE Questions for Class 11 Engineering Maths Permutations And Combinations Quiz 6 - 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 6 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

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