Permutations And Combinations(11th Grade > Mathematics ) Questions and answers for exam Preparation

11th Grade > Mathematics

PERMUTATIONS AND COMBINATIONS MCQs

Total Questions : 30 | Page 3 of 3 pages

Question 21.

The number of ways in which 8 boys be seated at a round table so that two particular boys are next to each other is

8!2!
7!2!
6!2!
6!

Discuss Question

Answer: Option C. -> 6!2!
:
C
The number of ways in which this can be done = 6! 2!

Question 22.

The number of ways of arranging 6 players to throw the hand ball so that the oldest player may not throw first is

Discuss Question

Answer: Option B. -> 600
:
B

The number of ways in which this can be done = 6! – 5! = 600

Question 23.

If m parallel lines in plane are intersected by n parallel lines, then number of parallelograms formed is

$\frac{m! n!}{{(2!)}^{2}}$
$\frac{m! n!}{(m - 2)! (n - 2)!}$
$\frac{m! n!}{{(2!)}^{2} (m - 2)! (n - 2)!}$
$\frac{(m + n)!}{(m + n - 2)! 2!}$

Discuss Question

Answer: Option C. ->

\frac{m! n!}{{(2!)}^{2} (m - 2)! (n - 2)!}

:
C

$^{m} C_{2} .^{n} C_{2}$

Question 24.

The number of permutations that can be made out of the letters of the word “EQUATION” which start with a consonant and end with a consonant is

$2! 6!$
$3! 6!$
$3! 5!$
$2! 5!$

Discuss Question

Answer: Option B. ->

3! 6!

:
B
Consonants occupy 2 ends in

^{3} P_{2}

ways remaining 6 letters occupy 6 places in 6! Ways
So the required number of arrangements =

^{3} P_{2} . 6! = 3! 6!

Question 25.

m men and n women are to be seated in a row so that no two women sit together. If m > n, then the number of ways in which they can be seated is

$m! n!$
$m!^{m} P_{n}$
$n!^{m} P_{n}$
$m!^{m + 1} P_{n}$

Discuss Question

Answer: Option D. ->

m!^{m + 1} P_{n}

:
D
The number of ways in which they can be seated =

m! .^{m + 1} P_{n}

Question 26.

The number of ways that a volley ball 6 can be selected out of 10 players so that 2 particular players are excluded is

Discuss Question

Answer: Option D. -> 28
:
D
The number of ways selecting 6 out of 10 so that 2 particular players are always excluded is

^{10 - 2} C_{6}

Question 27.

The number of nine digit numbers that can be formed with different digits is

$9.8!$
$8.9!$
$9.9!$
$10!$

Discuss Question

Answer: Option C. ->

9.9!

:
C

Required number numbers = total - the number of numbers begining with 0 = $10! - 9! = 9.9!$

Question 28.

The number of four digit even numbers that can be formed with 0,1,2,3,7,8, is

Discuss Question

Answer: Option D. -> 156
:
D

If 0 is in units place no. of ways = $^{5} P_{3} = 60$
If 2 or 8 is in units place no. of ways = $2 (^{5} P_{3} -^{4} P_{2}) = 96$
Total : 60 + 96 = 156

Question 29.

The number of permutations of n dissimilar things taken not more than ‘r’ at a time, when each thing may occur any number of times is

$\frac{n (n^{r} - 1)}{n - 1)}$
$\frac{n (n^{n} - n^{r})}{n - 1)}$
$^{n} P_{1} +^{n} P_{2} + \dots \dots +^{n} P_{r}$
$\frac{n {(n - 1)}^{r}}{n - 1}$

Discuss Question

Answer: Option A. ->

\frac{n (n^{r} - 1)}{n - 1)}

:
A

n + n^{2} + \dots \dots \dots + n^{r} = \frac{n (n^{r} - 1)}{n - 1}

Question 30.

Total 4 digit odd numbers that can be formed, if the digits used is not to be repeated again is

2240
2420
2440
2520

Discuss Question

Answer: Option A. -> 2240
:
A
The number of four digit numbers which satisfy the above condition =

8 \times 8 \times 7 \times 5 = 2240

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