Permutations & Combinations Questions and Answers

Answer: A

ENCYCLOPAEDIA has thirteen letters: vowels E, O, A, E, A, I (six, with E and A each twice) and consonants N, C, Y, C, L, P, D (seven, with C twice).

Positions 2, 4, 6, 8, 10 and 12 are the six even ones - exactly enough for the vowels.

\[ \text{vowels} = \frac{6!}{2!\,2!} = \frac{720}{4} = 180 \]

\[ \text{consonants in the 7 odd positions} = \frac{7!}{2!} = 2520 \]

\[ 180 \times 2520 = 453{,}600 \]

The counts of vowels and even positions match exactly, which is what makes the arrangement possible at all.

Answer: C

Each question has five choices, answered independently.

\[ 5^{4} = 625 \text{ ways to answer the paper} \]

Exactly one of those has all four right, so

\[ 625 - 1 = 624 \]

The complement is the whole trick: counting the ways to get at least one wrong directly would need four separate cases.

Answer: C

Count all sequences and remove the forbidden ones.

\[ \text{total} = 7! = 5040 \]

\[ \text{youngest last} = 6! = 720 \]

\[ 5040 - 720 = 4320 \]

Equivalently, the last place has 6 acceptable players and the rest arrange freely: \( 6 \times 6! = 4320 \).

Answer: B

The two selections are independent, so their counts multiply.

\[ {}^{5}C_{2} \times {}^{6}C_{3} = 10 \times 20 = 200 \]

Order does not matter within either group, so combinations are used rather than permutations.

Answer: B

The claim is false. Circular permutations of \( n \) distinct objects number

\[ (n-1)! \]

not \( n! \). In a circle there is no fixed starting point, so the \( n \) rotations of any arrangement all look the same - fixing one object's position removes that redundancy and leaves \( (n-1)! \).

Answer: C

MARKER has six letters with R twice, and its vowels are A and E.

\[ \text{total arrangements} = \frac{6!}{2!} = 360 \]

Treating the two vowels as one block gives five units:

\[ \frac{5!}{2!} \times 2! = 60 \times 2 = 120 \text{ with the vowels together} \]

\[ 360 - 120 = 240 \]

The division by \( 2! \) throughout accounts for the identical R's.

Answer: C

Count all pairs, then remove those that break the rule.

\[ \text{total} = {}^{5}C_{2} = 10 \]

Invalid pairs are those containing Mathematics 2 without Mathematics 1 - that is, Mathematics 2 paired with commerce, economics or statistics:

\[ 3 \text{ invalid pairs} \]

\[ 10 - 3 = 7 \]

The pair {Mathematics 1, Mathematics 2} is allowed, since the prerequisite is satisfied within the pair itself.

Answer: D

ALIVE has five letters, all different.

\[ 5! = 120 \]

No letter repeats, so there is no division to make - every arrangement is distinct.

Answer: B

The two choices are independent.

\[ 20 \times 25 = 500 \]

Every one of the 20 boys can be paired with every one of the 25 girls.

Answer: C

There are \( 2 + 3 + 4 = 9 \) chairs. "At least one yellow" is easiest through the complement.

\[ \text{total} = {}^{9}C_{3} = 84 \]

\[ \text{no yellow} = {}^{6}C_{3} = 20 \]

\[ 84 - 20 = 64 \]

Counting exactly one, two and three yellows separately gives \( 3 \times 15 + 3 \times 6 + 1 = 64 \) as well - the same answer for three times the work.