Permutations & Combinations Questions and Answers

Answer: C

This type of question becomes very easy when we assume three colour are red(R) blue(B) and Green(G). 

Now according to the statement 1 i.e.., codes can be generated by waving single flag of different colours, then number of ways are three i.e.., R.B.G from statement III three flags in different sequence of colours, then number of ways are six i.e.., RBG, BGR, GBR, RGB, BRG, GRB. 

Hence total number of ways by changing flag = 3+ 6 +6 = 15

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

Count all pairs, then remove those that break the rule.

\[ \text{total} = {}^{5}C_{2} = 10 \]

Invalid pairs are those containing Mathematics 2 without Mathematics 1 - that is, Mathematics 2 paired with commerce, economics or statistics:

\[ 3 \text{ invalid pairs} \]

\[ 10 - 3 = 7 \]

The pair {Mathematics 1, Mathematics 2} is allowed, since the prerequisite is satisfied within the pair itself.

Answer: D

Choose the principal first, since that post is the restricted one.

\[ \text{principal} = 2 \text{ ways} \]

The two vice-principals then come from the remaining five candidates:

\[ {}^{5}C_{2} = 10 \]

\[ 2 \times 10 = 20 \]

20 is not among the choices, so the answer is none of these. The two vice-presidencies are interchangeable, which is why a combination is used for them and not a permutation.

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

Deal with the most restricted assignment first.

Task 2 must go to person 3 or person 4:

\[ 2 \text{ ways} \]

Task 1 then cannot go to person 1 or 2, and one of persons 3 and 4 is already used, so two people remain eligible for it:

\[ 2 \text{ ways} \]

The other three tasks go to the remaining three people freely:

\[ 3! = 6 \]

\[ 2 \times 2 \times 6 = 24 \]

Assigning the constrained tasks first keeps the later counts independent of which choice was made.

Answer: C

For each visualization let us name the females and Females(3) Males(5) 

A B C D E F G H 

Since A cannot go with B. 

She will make team with four males in four ways AD,AF,AG,AH.

Since there is no compulsion with females C and E. 

They can make team with 5 males in 5 different ways each. 

Therefore, Total number of ways = 4+5+5 =14

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: 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: D

Count all arrangements, then remove those with a same-coloured pair adjacent.

\[ \text{total} = \frac{6!}{2!\,2!\,2!} = \frac{720}{8} = 90 \]

Treating one colour's pair as a single block:

\[ \frac{5!}{2!\,2!} = 30 \quad \text{for each of the 3 colours} \]

Two colours blocked:

\[ \frac{4!}{2!} = 12 \quad \text{for each of the 3 pairs of colours} \]

All three blocked: \( 3! = 6 \).

By inclusion–exclusion the number with at least one adjacent pair is

\[ 3(30) - 3(12) + 6 = 90 - 36 + 6 = 60 \]

\[ 90 - 60 = 30 \]

All three correction terms are needed; stopping after the first subtraction would leave zero.