Permutations & Combinations Questions and Answers

Q: From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there on the committee. In how many ways can it be done?

"At least 3 men" among five people splits into three cases.\[ 3M, 2W: \ {}^{7}C_{3} \times {}^{6}C_{2} = 35 \times 15 = 525 \]\[ 4M, 1W: \ {}^{7}C_{4} \times {}^{6}C_{1} = 35 \times 6 = 210 \]\[ 5M, 0W: \ {}^{7}C_{5} = 21 \]\[ 525 + 210 + 21 = 756 \]The cases are mutually exclusive, so the counts add. Here the complement would need just as much work, since "fewer than 3 men" also has three cases.

Q: In how many ways can the letters of the word LEADER be arranged ?

Correct answer: 360The word has 6 letters, of which E occurs twice and the rest once each.\[\frac{6!}{2!} = \frac{720}{2} = 360\]Dividing by 2! removes the arrangements that merely swap the two identical E’s.

Q: In how many ways can the letters of the word 'LEADER' be arranged?

LEADER has six letters with E appearing twice.\[ \frac{6!}{2!} = \frac{720}{2} = 360 \]The division by \( 2! \) removes the double-counting from swapping the two identical E's, which changes nothing visible.

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71.

From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there on the committee. In how many ways can it be done?

A.564
B.645
C.735
D.756

Answer: D

"At least 3 men" among five people splits into three cases.

\[ 3M, 2W: \ {}^{7}C_{3} \times {}^{6}C_{2} = 35 \times 15 = 525 \]

\[ 4M, 1W: \ {}^{7}C_{4} \times {}^{6}C_{1} = 35 \times 6 = 210 \]

\[ 5M, 0W: \ {}^{7}C_{5} = 21 \]

\[ 525 + 210 + 21 = 756 \]

The cases are mutually exclusive, so the counts add. Here the complement would need just as much work, since "fewer than 3 men" also has three cases.

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72.

In how many ways can the letters of the word INDIA be arranged, such that all vowels are never together?

A.48
B.42
C.28
D.36

Answer: B

INDIA has five letters with I twice, and its vowels are I, I, A.

\[ \text{total} = \frac{5!}{2!} = 60 \]

With the three vowels as one block there are three units (block, N, D):

\[ 3! \times \frac{3!}{2!} = 6 \times 3 = 18 \text{ with the vowels together} \]

\[ 60 - 18 = 42 \]

The repeated I forces a division by \( 2! \) in both counts.

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73.

There are 20 points in a plane, how many triangles can be formed by these points if 5 are colinear?

A.1130
B.550
C.1129
D.1140

Answer: A

A triangle needs three points, but three collinear points give no triangle.

\[ \text{all triples} = {}^{20}C_{3} = 1140 \]

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

\[ 1140 - 10 = 1130 \]

Only the five collinear points can produce a degenerate triple, so just one subtraction is needed.

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74.

A person ordered 5 pairs of black socks and some pairs of brown socks. The price of a black pair was thrice that of a brown pair while preparing the bill the bill clerk inter-changed the number of black and brown pairs by mistake which increased the bill by 100% what was the number of pair of brown socks in the original order?

A.10
B.15
C.20
D.25

Answer: D

Let the bought x pairs of brown socks and the price of each brown pair be Y.

Then total cost = 5x3Y+xy

Changed cost = 5xY+x*3Y

According to the question .

5y+3xy -(15y+xy)/15y+xy x 100% = 100%

=> 5y+3xy-15y-xy/ 15y+xy x100% = 100%

=> 5y+3xy = 15y+xy+15y+xy

=> 5y+3xy = 30y+2xy

=> xy= 25y

=> x=25.

Hence the original pair of brown socks = 25.

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75.

In a society, it is customary for friends of the same sex to hug and for friends of opposite sex to shake hands when they meet. A group of friends met in a party and there were 24 hand shakers. Which one among the following numbers indicates the possible number of hugs?

A.39
B.30
C.21
D.20

Answer: C

Let there be \( m \) men and \( f \) women. Handshakes happen only between opposite sexes and hugs only within a sex.

\[ mf = 24, \qquad \text{hugs} = {}^{m}C_{2} + {}^{f}C_{2} \]

Testing the factor pairs of 24:

\( 2 \times 12 \): \( 1 + 66 = 67 \)
\( 3 \times 8 \): \( 3 + 28 = 31 \)
\( 4 \times 6 \): \( 6 + 15 = 21 \)

So 21 is attainable, with four of one sex and six of the other.

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76.

In how many ways can the letters of the word LEADER be arranged ?

A.72
B.144
C.360
D.720

Answer: C

Correct answer: 360

The word has 6 letters, of which E occurs twice and the rest once each.

\[\frac{6!}{2!} = \frac{720}{2} = 360\]

Dividing by 2! removes the arrangements that merely swap the two identical E’s.

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77.

A bank has 6 digit account number with no repetition of digits within a account number. The first and last digit of the account numbers is fixed to be 4 and 7. How many such account numbers are possible?

A.10080
B.5040
C.890
D.1680

Answer: D

The first and last digits are fixed at 4 and 7, so only the four middle positions are free - and they must avoid 4 and 7, leaving eight digits.

\[ 8 \times 7 \times 6 \times 5 = 1680 \]

Repetition is forbidden, so each position has one fewer choice than the last.

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78.

Twelve people from a club, by picking lots. One of them will host a dinner for all once in a month. The number of dinners a particular member has to host in one year is

A.One
B.Zero
C.Three
D.Cannot be determined

Answer: D

Each month's host is drawn by lot from all twelve members, so the draws are independent and a member's chance does not depend on whether they have hosted before.

Over twelve months a particular member might be drawn every time, never, or anything in between - the count is not fixed.

So the answer cannot be determined. It would be exactly one only if the lots were drawn without replacement.

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79.

In how many ways can the letters of the word 'LEADER' be arranged?

A.72
B.144
C.360
D.720

Answer: C

LEADER has six letters with E appearing twice.

\[ \frac{6!}{2!} = \frac{720}{2} = 360 \]

The division by \( 2! \) removes the double-counting from swapping the two identical E's, which changes nothing visible.

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80.

A letter lock consists of three rings each marked with six different letters. The number of distinct unsuccessful attempts to open the lock is at the most -.

A.216
B.243
C.215
D.729

Answer: C

Each of the three rings has six letters, and they turn independently.

\[ 6 \times 6 \times 6 = 216 \text{ total settings} \]

Exactly one setting opens the lock, so the unsuccessful attempts number at most

\[ 216 - 1 = 215 \]

The "at most" matters: in the worst case every wrong setting is tried before the right one.

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