Permutations and Combinations — Practice

Question 1

A restaurant offers 4 starters, 5 main courses, and 3 desserts. In how many ways can a customer choose one of each?

Accepted Answer

By the multiplication principle, since the three choices are made independently: 4 x 5 x 3 = 60. Answer: There are 60 possible meal combinations.

Question 2

Evaluate 8!/(6! x 2!).

Accepted Answer

8! = 8x7x6!, so 8!/6! = 8x7 = 56. Dividing by 2!=2: 56/2 = 28. Answer: The value is 28.

Question 3

How many 3-digit numbers can be formed using the digits 1 to 9 if no digit is repeated?

Accepted Answer

This counts permutations of 9 distinct digits taken 3 at a time: 9P3 = 9!/6! = 9x8x7 = 504. Answer: There are 504 such 3-digit numbers.

Question 4

How many 4-letter codes can be formed using the 26 letters of the alphabet if repetition of letters is allowed?

Accepted Answer

Each of the 4 positions independently has 26 choices, since repetition is allowed: 26^4 = 456976. Answer: There are 456976 possible codes.

Question 5

Find the number of arrangements of the letters of the word STATISTICS.

Accepted Answer

STATISTICS has 10 letters: S appears 3 times, T appears 3 times, I appears 2 times, A and C once each. Arrangements = 10!/(3!3!2!) = 3628800/72 = 50400. Answer: There are 50400 distinct arrangements.

Question 6

How many arrangements of the letters of the word ORANGE are there in which the vowels (O, A, E) always occur together?

Accepted Answer

ORANGE has 6 distinct letters with 3 vowels (O,A,E) and 3 consonants (R,N,G). Treating the vowels as one block gives 4 objects to arrange (4!), and the vowels internally rearrange in 3! ways: 4! x 3! = 24 x 6 = 144. Answer: There are 144 such arrangements.

Question 7

In how many ways can a team of 3 be selected from 8 people?

Accepted Answer

Order does not matter for a team, so this is a combination: 8C3 = 8!/(3!5!) = (8x7x6)/(3x2x1) = 56. Answer: There are 56 possible teams.

Question 8

How many chords can be drawn through 15 points lying on a circle?

Accepted Answer

Each chord corresponds to an unordered choice of 2 of the 15 points: 15C2 = (15x14)/2 = 105. Answer: There are 105 possible chords.

Question 9

Verify Pascal's identity for n=5, r=2, i.e. confirm that 5C1 + 5C2 = 6C2.

Accepted Answer

5C1 = 5. 5C2 = 10. Sum = 15. Separately, 6C2 = (6x5)/2 = 15. Answer: Both sides equal 15, confirming 5C1+5C2=6C2.

Question 10

A committee of 4 is to be selected from 5 men and 6 women, with at most 1 man on the committee. Find the number of ways.

Accepted Answer

At most 1 man means 0 men or 1 man. 0 men, 4 women: 5C0 x 6C4 = 1 x 15 = 15. 1 man, 3 women: 5C1 x 6C3 = 5 x 20 = 100. Total = 15+100 = 115. Answer: There are 115 possible committees.

Practice — Permutations and Combinations

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