Permutations & Combinations Questions and Answers

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

Answer: C

One particular man and one particular woman are fixed in the group, so only the rest are free to choose.

\[ \text{2 more men from the remaining 4} = {}^{4}C_{2} = 6 \]

\[ \text{1 more woman from the remaining 5} = {}^{5}C_{1} = 5 \]

\[ 6 \times 5 = 30 \]

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

Answer: C

For each problem the candidate has four options: attempt one of the three internal choices, or leave the problem alone.

\[ 4^{5} = 1024 \]

One of those is attempting nothing at all, which the question excludes:

\[ 1024 - 1 = 1023 \]

The 511 on offer comes from \( 2^{9} - 1 \) and ignores the internal choices.

Answer: A

The two cases are mutually exclusive, so their counts add.

3 teachers and 3 boys:

\[ {}^{5}C_{3} \times {}^{4}C_{3} = 10 \times 4 = 40 \]

2 girls and 4 teachers:

\[ {}^{3}C_{2} \times {}^{5}C_{4} = 3 \times 5 = 15 \]

\[ 40 + 15 = 55 \]

Both cases give exactly six members, as required, and no group satisfies both conditions at once - so nothing is double-counted.

Answer: C

Treat the five boys as a single block, giving four units to arrange - the block and the three girls.

\[ 4! = 24 \]

Within the block the boys can be ordered in

\[ 5! = 120 \text{ ways} \]

\[ 24 \times 120 = 2880 \]

The block itself may sit anywhere among the girls, which is what the \( 4! \) counts.