Permutations And Combinations(12th Grade > Mathematics ) Questions and answers for exam Preparation

12th Grade > Mathematics

PERMUTATIONS AND COMBINATIONS MCQs

Permutations And Combinations

Total Questions : 60 | Page 6 of 6 pages

T_{n}

be the number of all possible triangles formed by joining vertices of an n-sided regular polygon. If

T_{n + 1} - T_{n} = 10

, then the value of N is

Discuss Question

Answer: Option B. -> 5
:
B
Given,

T_{n} =^{n} C_{3} \Rightarrow T_{n + 1} =^{n + 1} C_{3}

∴ T_{n + 1} - T_{n} =^{n + 1} C_{3} -^{n} C_{3} = 10

[given]

\Rightarrow^{n} C_{2} +^{n} C_{3} -^{n} C_{3} = 10

[∵^{n} C_{r} +^{n} C_{r + 1} =^{n + 1} C_{r + 1}]

\Rightarrow^{n} C_{2} = 10 \Rightarrow n = 5

Question 52. A debate club consists of 6 girls and 4 boys. A team of 4 members is to be selected from this club including the selection of a captain (from among these 4 members) for the team. If the team has to include atmost one boy, the number of ways of selecting the team is ?

Discuss Question

Answer: Option A. -> 380
:
A
We have, 6 girls and 4 boys. To select 4 members (atmost one boy)
i.e. (1 boy and 3 girls) or (4 girls)

=^{6} C_{3} .^{4} C_{1} +^{6} C_{4}

.....(i)
Now, selection of captain from 4 members (including the selection of a captain, from these 4 members)

(^{6} C_{3} .^{4} C_{1} +^{6} C_{4})^{4} C_{1} = (20 \times 4 + 15) \times 4 = 380

Question 53. 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 54. If m parallel lines in plane are intersected by n parallel lines, then number of parallelograms formed is

m!n!(2!)2
m!n!(m−2)!(n−2)!
m!n!(2!)2(m−2)!(n−2)!
(m+n)!(m+n−2)!2!

Discuss Question

Answer: Option C. -> m!n!(2!)2(m−2)!(n−2)!
:
C

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

Question 55. 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

n(nr−1)n−1)
n(nn−nr)n−1)
nP1+nP2+⋯⋯+nPr
n(n−1)rn−1

Discuss Question

Answer: Option A. -> n(nr−1)n−1)
:
A

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

Question 56. 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

Question 57. 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