Permutations & Combinations Questions and Answers

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.

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

There are \( 2 + 3 + 5 = 10 \) balls, 7 of them not black. Use the complement.

\[ \text{total} = {}^{10}C_{3} = 120 \]

\[ \text{no black} = {}^{7}C_{3} = 35 \]

\[ 120 - 35 = 85 \]

Counting one, two and three black balls separately gives \( 3 \times 21 + 3 \times 7 + 1 = 85 \) as well.

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: D

There are ten children. "At least one boy" is easiest through the complement.

\[ \text{total} = {}^{10}C_{4} = 210 \]

\[ \text{no boy} = {}^{4}C_{4} = 1 \]

\[ 210 - 1 = 209 \]

With only four girls there is exactly one all-girl selection, so the complement is a single case.

Answer: C

ALIVE has five distinct letters.

\[ 5! = 120 \text{ arrangements in all} \]

One of those is the original word itself, which is not new:

\[ 120 - 1 = 119 \]

The 120 on offer is the total count and 24 is \( 4! \) - both a step away from what is asked.

Answer: D

A number is even only if its last digit is even, and 2 is the only even digit available.

\[ \text{last digit} = 1 \text{ choice} \]

The other three positions take the remaining five digits without repetition:

\[ 5 \times 4 \times 3 = 60 \]

\[ 1 \times 60 = 60 \]

Fixing the constrained position first is the standard tactic for these.

Answer: A

The two particular ladies are already in, so only three more places remain and they come from the other six people (1 lady and 5 gentlemen).

\[ {}^{6}C_{3} = 20 \]

Fixing members reduces both the size of the pool and the number still to be chosen.

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: B

Case I if lady sits on the reserved seat, then 2 men can occupy seats from 4 vacant seats in = 4P2 = 4x3 = 12ways 

Case II if lady does not sit on reversed seat, then I. Woman can occupy a seat from four seats in 4 ways. I. man can occupy a seat from 3 seats in 3 ways, also I. man left can occupy a seat from remaining two seats in 2 ways. 

Therefore, Total ways = 4x3x2 = 24ways 

From case I and case II 

Total number of ways = 12+24 = 36