This set of Discrete Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “Counting – Circular Permutations”.

1. In a playground, 3 sisters and 8 other girls are playing together. In a particular game, how many ways can all the girls be seated in a circular order so that the three sisters are not seated together?

a) 457993

b) 3386880

c) 6544873

d) 56549

View Answer

Explanation: There are 3 sisters and 8 other girls in total of 11 girls. The number of ways to arrange these 11 girls in a circular manner = (11– 1)! = 10!. These three sisters can now rearrange themselves in 3! ways. By the multiplication theorem, the number of ways so that 3 sisters always come together in the arrangement = 8! × 3!. Hence, the required number of ways in which the arrangement can take place if none of the 3 sisters is seated together: 10! – (8! × 3!) = 3628800 – (40320 * 6) = 3628800 – 241920 = 3386880.

2. How many numbers of three digits can be formed with digits 1, 3, 5, 7 and 9?

a) 983

b) 120

c) 345

d) 5430

View Answer

Explanation: Here number of digits, n = 5 and number of places to be filled-up r = 3. Hence, the required number is

^{5}P

_{3}= 5!/2!*3! = 120.

3. The size of a multiset is 6 which is equal to the number of elements in it with counting repetitions (a multiset is an unordered collection of elements where the elements may repeat any number of times). Determine the number of multisets can be grouped from n distinct elements so that at least one element occurs exactly twice?

a) 326

b) 28

c) 45

d) 62

View Answer

Explanation: There are six places to be filled in the multiset using the n distinct elements. At least one element has to occur exactly twice and that would leave 4 more places in the multiset means that at most four elements can occur exactly once. Thus there are two mutually exclusive cases as follows: 1) Exactly one element occurs exactly twice and select this element in n ways. Fill up the remaining four spots using 5 distinct elements from the remaining n−1 elements in

^{n-1}C

_{4}ways. 2) Exactly four elements that occur at least once each. Hence, the total number of ways to form the multiset is

^{n}C

_{2}+ n *

^{n-1}C

_{4}=

^{6}C

_{2}+ 6 *

^{6-1}C

_{4}= 45.

4. How many words can be formed with the letters of the word ‘CASTLE’ when ‘O’ and ‘A’ occupying end places.

a) 217

b) 48

c) 75

d) 186

View Answer

Explanation: When ‘O’ and ‘A’ are occupying end-places => A.S.T.L. (CE). We can see that (CE) are fixed, hence A, S, T, L can be arranged in 4! Ways and (C, E) can be arranged themselves is 2! ways. So, the number of words formed = 4! x 2! = 48 ways.

5. Determine the number of ways of choosing a cricket team (consists of 11 players) out of 18 players if a particular player is never chosen.

a) 12798

b) 22800

c) 31824

d) 43290

View Answer

Explanation: If a particular player is never chosen that would mean 11 players are selected out of 18 players. Hence, required number of ways =

^{18}C

_{11}= 31824.

6. How many different choices can be made from 5 roses, 4 marigold and 8 sunflowers if at least one flower is to be chosen for making of garland?

a) 269

b) 270

c) 281

d) 320

View Answer

Explanation: Number of ways of selecting roses = (5+1) = 6 ways, number of ways of selecting marigold = (4+1) = 5 ways, and the number of ways of selecting sunflowers = (8+1) = 9 ways. Total number of ways of selecting flowers= 6 * 5 * 9 = 270. But this includes when no flowers or zero flowers is selected (There is no flowers of a different type, hence n=0 => 2

^{n}= 2

^{0}= 1). Hence, the number of ways of selecting at least one fruit = 270 – 1 = 269.

7. In how many ways 6 pens can be selected from 15 identical black pens?

a) 9*3!

b) 21

c) 14!

d) 1

View Answer

Explanation: Here the pens are identical, the total number of ways of selecting 6 pens is 1.

8. Determine the number of ways of selecting one or more letters from the letters BBBBBB?

a) 6

b) 73

c) 23

d) 56

View Answer

Explanation: The number of ways of selecting one ‘B’s = 1, selecting two ‘B’s = 1, selecting three ‘B’s = 1, selecting four ‘B’s = 1, selecting five ‘B’s = 1 and selecting six ‘B’s = 1. Hence, the required number of ways = 6.

9. Determine the number of ways such that 5 men and 5 women be seated at a round table if no two women are seated together.

a) 654870

b) 144521

c) 2880

d) 5634

View Answer

Explanation: We can arrange the 5 men around the table in (5-1)! = 4! ways (circular permutation). Then, we place 5 women in the empty spaces between the men, ensuring no two women sit together. This can be done by placing a woman in every other space, which can be achieved in 5! ways.

Therefore, the total number of arrangements is 4! * 5! = 2880.

10. Find the number of ways in which 4 people E, F, G, H, A, C can be seated at a round table, such that E and F must always sit together.

a) 32

b) 290

c) 124

d) 48

View Answer

Explanation: E and F can sit together in all arrangements in 2! Ways. Now, the arrangement of the 5 people in a circle can be done in (5 – 1)! or 24 ways. Therefore, the total number of ways will be 24 x 2 = 48.

**Sanfoundry Global Education & Learning Series – Discrete Mathematics.**

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