Permutations & Combinations Questions and Answers

Q: The number of new words that can be formed by rearranging the letters of the word 'ALIVE' is -.

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.

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31.

In a room there are 2 green chairs, 3 yellow chairs and 4 blue chairs. In how many ways can Raj choose 3 chairs so that at least one yellow chair is included?

A.3
B.30
C.64
D.84

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.

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32.

In how many different ways can the letters of the word 'LEADING' be arranged in such a way that the vowels always come together?

A.360
B.480
C.720

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.

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33.

A box contains 2 white, 3 black and 5 red balls. In how many ways can three balls be drawn from the box if at least one black ball is to be included in the draw?

A.29
B.36
C.48
D.85

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.

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34.

In how many different ways can six players be arranged in a line such that two of them, Ajeet and Mukherjee are never together?

A.120
B.240
C.360
D.480

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.

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35.

In a group of 6 boys and 4 girls, four children are to be selected. In how many different ways can they be selected such that at least one boy should be there?

A.159
B.194
C.205
D.209

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.

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36.

The number of new words that can be formed by rearranging the letters of the word 'ALIVE' is -.

A.24
B.23
C.119
D.120

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.

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37.

How many four digit even numbers can be formed using the digits {2, 3, 5, 1, 7, 9}

A.100
B.360
C.120
D.60

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.

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38.

A delegation of 5 members has to be formed from 3 ladies and 5 gentlemen. In how many ways the delegation can be formed, if 2 particular ladies are always included in the delegation?

A.20
B.54
C.42
D.60

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.

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39.

A mixed doubles tennis game is to be played two teams(each consists of one male and one female) There are four married couples. No team is to consist a husband and his wife. What is the maximum number of games that can be played?

A.12
B.48
C.36
D.42

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.

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40.

2 men and 1 woman board a bus of which 5 seats are vacant, one of these 5 seats is reserved for ladies. A woman may or may not sit on the seat reserved for ladies, In how many different ways can the five seats be occupied by these passengers?

A.15
B.36
C.48
D.60

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

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