Permutations & Combinations Questions and Answers

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.