Permutations & Combinations Questions and Answers

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.

Answer: B

MEADOWS has seven distinct letters: vowels E, A, O and consonants M, D, W, S.

The even places among positions 1 to 7 are 2, 4 and 6 - three places for three vowels.

\[ \text{vowels} = 3! = 6 \]

\[ \text{consonants in the 4 odd places} = 4! = 24 \]

\[ 6 \times 24 = 144 \]

The counts match exactly, which is what makes the restriction satisfiable.

Answer: B

(i) Two particular members must be included. They take two of the four places, so two more come from the remaining six:

\[ {}^{6}C_{2} = 15 \]

(ii) Two particular members must not be included. All four come from the remaining six:

\[ {}^{6}C_{4} = 15 \]

The two answers coincide because \( {}^{6}C_{2} = {}^{6}C_{4} \) - choosing two to include is the same as choosing four to leave out.

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.