Permutations & Combinations Questions and Answers

Answer: C

Each of the three rings has six letters, and they turn independently.

\[ 6 \times 6 \times 6 = 216 \text{ total settings} \]

Exactly one setting opens the lock, so the unsuccessful attempts number at most

\[ 216 - 1 = 215 \]

The "at most" matters: in the worst case every wrong setting is tried before the right one.

Answer: D

SMOKEJACK has nine letters with K appearing twice. J must be one of the four, so three more are chosen from S, M, O, K, E, A, C - with K available twice.

At most one K - choose 3 of the 7 distinct letters:

\[ {}^{7}C_{3} = 35 \]

Both K's - choose 1 more from the other 6:

\[ {}^{6}C_{1} = 6 \]

\[ 35 + 6 = 41 \]

The repeated K is what makes this more than a plain \( {}^{7}C_{3} \); the two-K selections are genuinely different combinations.

Answer: A

Seven modes are available going out, and the return must differ from the one used.

\[ 7 \times 6 = 42 \]

The restriction removes exactly the seven journeys that would repeat a mode.

Answer: D

Choose the principal first, since that post is the restricted one.

\[ \text{principal} = 2 \text{ ways} \]

The two vice-principals then come from the remaining five candidates:

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

\[ 2 \times 10 = 20 \]

20 is not among the choices, so the answer is none of these. The two vice-presidencies are interchangeable, which is why a combination is used for them and not a permutation.

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: B

The word 'OPTICAL' contains 7 different letters.

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

Then, we have to arrange the letters PTCL (OIA).

Now, 5 letters can be arranged in 5! = 120 ways.

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

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

Answer: D

The two choices are independent, so they multiply.

\[ 9 \times 12 = 108 \]

Each of the nine trousers can be paired with each of the twelve shirts. Adding them would answer a different question - how many garments he could pick one of.

Answer: B

The box holds \( 4 + 3 + 6 = 13 \) marbles.

\[ \text{total ways} = {}^{13}C_{2} = 78 \]

\[ \text{same colour} = {}^{4}C_{2} + {}^{3}C_{2} + {}^{6}C_{2} = 6 + 3 + 15 = 24 \]

\[ P = \frac{24}{78} \]

The three colours are mutually exclusive, so their counts add before dividing.

Answer: B

CORPORATION has eleven letters, but repetition of letters is not allowed in the word formed, so only the distinct letters may be used:

\[ C,\ O,\ R,\ P,\ A,\ T,\ I,\ N \rightarrow 8 \text{ distinct} \]

Order matters in a word, so this is a permutation:

\[ {}^{8}P_{3} = 8 \times 7 \times 6 = 336 \]

The repeated O's and R's contribute nothing extra, since no letter may be used twice.

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.