Permutations & Combinations Questions and Answers

Answer: B

Count all orders, then remove those with A and B together.

\[ \text{total} = 4! = 24 \]

Treating A and B as one block gives three units, and the pair can be either way round:

\[ 3! \times 2! = 6 \times 2 = 12 \]

\[ 24 - 12 = 12 \]

With only four books, exactly half the arrangements keep A and B apart.

Answer: C

First player can post greeting cards to the remaining 54 players in 54 ways. Second player can post greeting card to the 54 players. Similarly, it happens with the rest of the players. The total numbers of greeting cards posted are
54 + 54 + 54 …
54 (55times) = 54 x 55 = 2970.

Answer: C

The given word is DANGER. Number of letters is 6. Number of vowels is 2 (i.e., A, E). Number of consonants is 4 (i.e., D,N,G,R). As the vowels cannot occupy odd places, they can be arranged in even places. Two vowels can be arranged in 3 even places in 3P2 ways i.e., 6. Rest of the consonants can arrange in the remaining 4 places in 4! ways. The total number of arrangements is 6 x 4! = 144.

Answer: B

Four-digit numbers from five distinct digits, no repetition:

\[ {}^{5}P_{4} = 120 \text{ numbers} \]

By symmetry each digit occupies each position equally often:

\[ \frac{120}{5} = 24 \text{ times per position} \]

The digits sum to \( 1+3+5+7+9 = 25 \), so each position contributes \( 24 \times 25 = 600 \) to its place value.

\[ \text{sum} = 600 \times (1000 + 100 + 10 + 1) = 600 \times 1111 \]

\[ = 666{,}600 \]

The symmetry argument avoids listing 120 numbers: no digit is favoured in any column.

Answer: B

A rectangle on the board is fixed by choosing two of the nine vertical lines and two of the nine horizontal ones.

\[ {}^{9}C_{2} \times {}^{9}C_{2} = 36 \times 36 = 1296 \]

The squares are counted by size: an \( n \times n \) square can sit in \( (9-n)^{2} \) positions.

\[ 8^{2} + 7^{2} + \cdots + 1^{2} = 204 \]

\[ \text{rectangles that are not squares} = 1296 - 204 = 1092 \]

So the answer is 1092 and 204. Every square is a rectangle, which is why the two counts must be separated before comparing.

Answer: A

A triangle needs three points, but three collinear points give no triangle.

\[ \text{all triples} = {}^{20}C_{3} = 1140 \]

\[ \text{collinear triples} = {}^{5}C_{3} = 10 \]

\[ 1140 - 10 = 1130 \]

Only the five collinear points can produce a degenerate triple, so just one subtraction is needed.

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: D

"At least 3 men" among five people splits into three cases.

\[ 3M, 2W: \ {}^{7}C_{3} \times {}^{6}C_{2} = 35 \times 15 = 525 \]

\[ 4M, 1W: \ {}^{7}C_{4} \times {}^{6}C_{1} = 35 \times 6 = 210 \]

\[ 5M, 0W: \ {}^{7}C_{5} = 21 \]

\[ 525 + 210 + 21 = 756 \]

The cases are mutually exclusive, so the counts add. Here the complement would need just as much work, since "fewer than 3 men" also has three cases.

Answer: C

The word 'LEADING' has 7 different letters.

When the vowels EAI are always together, they can be supposed to form one letter.

Then, we have to arrange the letters LNDG (EAI).

Now, 5 (4 + 1 = 5) letters can be arranged in 5! = 120 ways.

The vowels (EAI) can be arranged among themselves in 3! = 6 ways.

 Required number of ways = (120 x 6) = 720.

Answer: D

CORPORATION has eleven letters - O three times, R twice - with vowels O, O, A, I, O and consonants C, R, P, R, T, N.

Treating the five vowels as one block gives seven units, among which R repeats:

\[ \frac{7!}{2!} = 2520 \]

Within the block, the five vowels include three O's:

\[ \frac{5!}{3!} = 20 \]

\[ 2520 \times 20 = 50{,}400 \]

Both repetitions have to be divided out - the R's among the units and the O's inside the block.