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 1 of 6 pages

Question 1. The number of triangles that can be formed by choosing the vertices from a set of 12 points, seven of which lie on the same straight line is

Discuss Question

Answer: Option A. -> 185
:
A
Required number of ways =

^{12} C_{3} -^{7} C_{3}

= 220 - 35 = 185

Question 2. The number of numbers of 4 digits which are not divisible by 5 are

7200
3600
14400
1800

Discuss Question

Answer: Option A. -> 7200
:
A
The total number of 4 digits are 9999-999=9000.
The numbers of 4 digits number divisible by 5 are

90 \times 20

=1800. Hence required number of ways are 9000-1800 =7200.
{Since there are 20 numbers in each hundred (1 to 100) divisible by 5 and from 999 to 9999 there are 90 hundreds, hence the results}.

Question 3. The number of ways that a volley ball 6 can be selected out of 10 players so that 2 particular players are excluded is

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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 4. The number of diagonals in an octagon will be

Discuss Question

Answer: Option B. -> 20
:
B
Number of diagonals in an Octagon =

^{8} C_{2} - 8

= 20

Question 5. The number of ways in which the letters of the word ARRANGE can be arranged such that both R do not come together is

360
900
1260
1620

Discuss Question

Answer: Option B. -> 900
:
B
The word ARRANGE, has AA,RR, NGE letters. That is two A' s, two R's and N,G,E one each.

∴

The total number of arrangements

\frac{7!}{2! 2! 1! 1! 1!}

=1260
But, the number of arrangements in which bothRR are together as one unit =

\frac{6!}{2! 1! 1! 1! 1!}

= 360

∴

The number of arrangements in which both RR do not come together = 1260 -360 = 900.

Question 6. There are n straight lines in a plane, no two of which are parallel and no three pass through the same point. Their points of intersection are joined. Then the number of fresh lines thus obtained is

n(n−1)(n−2)8
n(n−1)(n−2)(n−3)6
n(n−1)(n−2)(n−3)8
n(n−1)(n−2)(n−3)4

Discuss Question

Answer: Option C. -> n(n−1)(n−2)(n−3)8
:
C
Since no two lines are parallel and no three are concurrent, therefore n straight lines intersect at

^{n} C_{2}

= N(say) points. Since two points are required to determine a straight line, therefore the total number of lines obtained by joining N points

^{N} C_{2}

. But in this each old line has been counted

^{n - 1} C_{2}

times, since on each old line there will be n-1 points of intersection made by the remaining (n-1) lines.
Hence the required number of fresh lines is

^{N} C_{2} - n .^{n - 1} C_{2}

\frac{N (N - 1)}{2} - \frac{n (n - 1) (n - 2)}{2}

\frac{^{n} C_{2} (^{n} C_{2} - 1)}{2} - \frac{n (n - 1) (n - 2)}{2}

\frac{n (n - 1) (n - 2) (n - 3)}{8}

Question 7. The value of

2^{n} {1.3.5..... (2 n - 3) (2 n - 1)}

(2n)!n!
(2n)!2n
n!(2n)!
(2n−1)!n!

Discuss Question

Answer: Option A. -> (2n)!n!
:
A

1.3.5...... (2 n - 1) 2^{n}

\frac{1.2.3.4.5.6.... (2 n - 1) (2 n) 2^{n}}{2.4.5.....2 n}

\frac{(2 n)! 2^{n}}{2^{n} (1.2.3...... n)}

\frac{(2 n)!}{n!}

Question 8. The number of diagonals in an octagon will be

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Answer: Option B. -> 20
:
B
Number of diagonals in an Octagon =

^{8} C_{2} - 8

= 20

Question 9. An auto mobile dealer provides motor cycles and scooters in 3 body patterns and 4 different colours each. The number of choices open to customer is

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Answer: Option D. -> 24
:
D
By fundamental theorem of Multiplication = 4

\times

\times

Question 10. The number of integers greater than 6000 that can be formed using the digits 3, 5, 6, 7 and 8 without repetition is

Discuss Question

Answer: Option B. -> 192
:
B
The integer greater than 6000 may be of 4 digits or 5 digits. So, here two cases arise.
Case I When number is of 4 digits.
Four digit number can start from 6, 7 or 8.