Permutations & Combinations Questions and Answers

Answer: B

MATERIAL has eight letters with A twice: vowels A, E, I, A and consonants M, T, R, L.

Treat the four vowels as one block, giving five units:

\[ 5! = 120 \]

Within the block the vowels arrange in

\[ \frac{4!}{2!} = 12 \text{ ways} \]

\[ 120 \times 12 = 1440 \]

The division by \( 2! \) accounts for the two identical A's; without it the count would double to 2880.

Answer: B

The three selections are independent, so their counts multiply.

\[ {}^{4}C_{2} \times {}^{6}C_{3} \times {}^{5}C_{1} \]

\[ = 6 \times 20 \times 5 = 600 \]

Answer: C

\[ {}^{4}C_{3} \times {}^{6}C_{2} \times {}^{5}C_{3} \]

\[ = 4 \times 15 \times 10 = 600 \]

Each group is chosen from its own pool, so the three counts are independent and multiply.

Answer: D

A line needs two points. Two points on the same line give nothing new - they just re-describe that line.

One point from each line:

\[ 6 \times 5 = 30 \text{ new lines} \]

The two original lines add 2 more.

\[ 30 + 2 = 32 \]

Taking \( {}^{11}C_{2} = 55 \) would count every pair, including the \( {}^{6}C_{2} + {}^{5}C_{2} = 25 \) collinear pairs that produce no new line - and \( 55 - 25 + 2 = 32 \) again.

Answer: D

Each of the ten questions is answered independently as T or F.

\[ 2^{10} = 1024 \]

Since no two candidates matched, at most 1024 candidates could sit the paper. The 512 on offer is \( 2^{9} \), one question short.

Answer: C

Count all groups of three, then remove the forbidden ones.

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

Groups containing both C and D need one more from A, B, E:

\[ {}^{3}C_{1} = 3 \]

\[ 10 - 3 = 7 \]

Counting the forbidden case and subtracting is quicker than enumerating the seven allowed groups directly.

Answer: D

As there are six players, So total ways in which they can be arranged = 6!ways =720.

A number of ways in which Ajeet and Mukherjee are together = 5!x2 = 240.

Therefore, Number of ways when they don’t remain together = 720 -240 =480.

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: D

Married couples = MF MF MF MF 

                              AB CD EF GH 

Possible teams = AD CB EB GB 

                             AF CF ED GD 

                             AH CH EH GF S

Since one male can be paired with 3 other female, Total teams = 4x3 = 12. 

Team AD can play only with CB,CF,CH,EB,EH,GB,GF(7 teams ) 

Team AD cannot play with AF, AH, ED and GD 

The same will apply with all teams, So number of total matches = 12x7 = 84. 

But every match includes 2 teams, so the actual number of matches = 84/2 = 42.

Answer: C

"At least one" is easiest through the complement.

\[ \text{total outcomes} = 6^{3} = 216 \]

\[ \text{no die shows 2} = 5^{3} = 125 \]

\[ 216 - 125 = 91 \]

Counting exactly one, exactly two and exactly three 2's gives \( 75 + 15 + 1 = 91 \) as well, for considerably more work.