Permutation Combination : Aptitude Questions with Answers

Ans: B) 60

Since “APPLE” has 5 letters, with the letter ‘P’ repeating twice, the number of distinct 5-letter words that can be formed is given by: $\frac{5!}{2!} = \frac{120}{2} = 60$

Understand the difference between applying permutations and combinations.

Ans: D

For F: Total combination is $50C_6$ = $\frac{504948474645}{654321}$

For G: Total combination is $50C_5$*$5C_1$ = $\frac{50*49*48*47*46}{5*4*3*2*1}$*$\frac{50}{1}$

% greater = $\frac{change}{smaller}$*100

change = $\frac{50*49*48*47*46}{5*4*3*2*1}$*$\frac{50}{1}$ – $\frac{50*49*48*47*46*45}{6*5*4*3*2*1}$

= $\frac{50*49*48*47*46}{5*4*3*2*1}$ (50 – $\frac{45}{6}$)

% greater = $\frac{\frac{50*49*48*47*46}{5*4*3*2*1}(50 – \frac{45}{6})}{\frac{50*49*48*47*46*45}{6*5*4*3*2*1}}$*100

= $\frac{50 – \frac{45}{6}}{\frac{45}{6}}$*100

= $\frac{50*6 – 45}{45}$*100

=566.66%

Ans: A) 453600

The word “INDEPENDENCE” has 12 letters: I, N (3 times), D (2 times), E (4 times), and C. So,

$\frac{12!}{3! \times 2! \times 4! \times 1!} = \frac{479001600}{6 \times 2 \times 24 \times 1} = 453600$

Ans: D) 286

This is a problem of distributing n identical items into r distinct groups, which is given by $\binom{n + r – 1}{r – 1}$.

$\binom{10 + 4 – 1}{4 – 1} = \binom{13}{3} = 286$

Ans: D) 3108

Case 1: 3 women, 3 men: (73)×(83)=$\binom{7}{3} \times \binom{8}{3} = 35 \times 56 = 1960$
Case 2: 4 women, 2 men: (74)×(82)=$\binom{7}{4} \times \binom{8}{2} = 35 \times 28 = 980$
Case 3: 5 women, 1 man: (75)×(81)=$\binom{7}{5} \times \binom{8}{1} = 21 \times 8 = 168$

$1960+980+168=3108$

Ans: C) 420

Treat all S’s as one unit. We then have: M, I, I, I, P, P, S block. The total number of arrangements is:

$\frac{7!}{3! \times 2!} = \frac{5040}{6 \times 2} = 420$

Ans: A) 28

If 2 particular people (the couple) are always included, we need to choose 2 more people from the remaining 8. This can be done in:

$\binom{8}{2} = 28$

Ans: B) 60480

Treat the vowels (A, A, E, E) as a single unit, giving us 8 units. Then,

$\frac{8!}{2! \times 2!} \times \frac{4!}{2! \times 2!} = 10080 \times 6 = 60480$

Ans: D) 1050

Consider cases for the number of women:

• 2 women, 3 men: $\binom{6}{2} \times \binom{7}{3} = 15 \times 35 = 525$
• 3 women, 2 men: $\binom{6}{3} \times \binom{7}{2} = 20 \times 21 = 420$
• 4 women, 1 man: $\binom{6}{4} \times \binom{7}{1} = 15 \times 7 = 105$

Total: $525+420+105=1050$

Ans: B) 1120

This is a problem of distributing identical items among distinct groups with a total number of ways:

$\binom{5+4-1}{4-1} \times \binom{3+4-1}{4-1} = \binom{8}{3} \times \binom{6}{3} = 56 \times 20 = 1120 $

Ans: C

There are $\binom{10}{2} = 45$ possible cases.

The formula for combinations is $\binom{n}{k} = \frac{n!}{k!(n-k)!}$, where $n$ is the total number of items and $k$ is the number of items being selected. In this case, $n=10$ and $k=2$, so the formula gives $\binom{10}{2} = \frac{10!}{2!(10-2)!} = \frac{10 \times 9}{2 \times 1} = 45$.

Ans: B

We have 7 total ducks, including one adult male, one adult female, and one ugly duckling. We need to determine the number of ways the ducks can line up with the adult ducks at each end and the ugly duckling in the center.

In the first spot, we have 2 choices (since it can be 1 of the 2 adult ducks).

In the second spot, we have 4 choices (since it can be any 1 of the 4 non-ugly ducklings).

In the third spot, we have 3 choices (since it can be any 1 of the 3 remaining non-ugly ducklings).

In the fourth spot, i.e., the center spot, we have only 1 choice (since it must be the ugly duckling).

In the fifth spot, we have 2 choices (since it can be any 1 of the 2 remaining non-ugly ducklings).

In the six spot, we have 1 choice (since it must be the last remaining non-ugly duckling).

In the seventh spot, i.e., the last position, we have 1 choice (since it must be the other adult duck).

Thus, the number of ways the ducks can lineup with an adult duck at each end and the ugly duckling in the middle is 2 x 4 x 3 x 1 x 2 x 1 x 1 = 48.

Ans: D

To solve this problem, we can first calculate the total number of unrestricted seating arrangements and then subtract the number of arrangements where Anya and Xavier are sitting next to each other.

1. **Total number of unrestricted arrangements:**

There are 5 people (Anya, Peter, Martha, Xavier, Ed). If there are no restrictions, any of the 5 people can sit in the first seat, any of the remaining 4 can sit in the second seat, and so on. This is a permutation of 5 people, calculated as:

$ 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 $

2. **Arrangements where Anya and Xavier are sitting next to each other:**

To simplify this, consider Anya and Xavier as a single unit. This effectively reduces the number of units to arrange from 5 to 4 (the unit “Anya-Xavier”, plus Peter, Martha, Ed).

– The unit “Anya-Xavier” can be arranged in 2 ways (Anya can be first or Xavier can be first).

– The total arrangement for the 4 units is $4!$, as there are now 4 units to arrange:

$ 4! = 4 \times 3 \times 2 \times 1 = 24 $

– Since the Anya-Xavier unit can be arranged in 2 ways, multiply their internal arrangements by the other possible arrangements of the units:

$ 2 \times 24 = 48 $

3. **Subtract the restricted arrangements from the total arrangements:**

Subtract the number of arrangements where Anya and Xavier are next to each other from the total number of unrestricted arrangements:

$ 120 – 48 = 72 $

Therefore, there are 72 different seating arrangements possible where Anya and Xavier are not sitting next to each other.

Ans: C

To find the total number of matches played in the first round of the Indian Premier League with each team playing every other team exactly twice, we follow these steps:

1. **Calculate the Number of Matches if Each Team Played Every Other Team Once:**

There are 8 teams, and each team plays every other team exactly once. The number of such matches can be calculated using the combination formula, which in this context is:

$ C(n, 2) = \frac{n(n-1)}{2} $

where $ n $ is the number of teams. Here, $ n = 8 $, so:

$ C(8, 2) = \frac{8 \times 7}{2} = 28 $

This means there would be 28 matches if each team played every other team exactly once.

2. **Each Team Plays Every Other Team Twice:**

Since each match scenario is played twice, the total number of matches will be:

$ 28 \times 2 = 56 $

Therefore, the total number of matches played in the first round with each team playing every other team exactly twice is 56.

Ans: C

There are 10 guests at the party, and Rondy does not want to stand next to Harinarayan. We can think of this problem as asking how many ways there are to arrange 10 people in a line if two of them (Harinarayan and Rondy) cannot be next to each other.

One way to solve this problem is to consider all 10! (10 factorial) possible arrangements of the 10 guests, and then subtract the number of arrangements where Harinarayan and Rondy are next to each other.

There are 9! (9 factorial) ways to arrange the 10 guests without considering Harinarayan and Rondy. For each of these arrangements, there are 2 ways to place Harinarayan and Rondy next to each other, so there are a total of 9! * 2 = 362880 arrangements where Harinarayan and Rondy are next to each other.

Therefore, there are 10! – 9! * 2 = 945,000 arrangements of the 10 guests where Harinarayan and Rondy cannot be next to each other.

Another way to solve this problem is to use the following formula:

Number of arrangements = n! / (r!(n – r)!)

where n is the number of guests and r is the number of guests who cannot be next to each other.

In this case, n = 10 and r = 2, so the number of arrangements is:

10! / (2!(10 – 2)!) = 945,000

Ans: A

The 3 red gumballs selected at the end are irrelevant, since they can be arranged only 1 way:

RRR

Thus, the question stem can be rephrased as follows:

Before the 3 red gumballs are selected at the end, in how many different ways can the other 7 gumballs be selected?

Number of ways to arrange 7 distinct elements = 7!

But the elements here are not all distinct; some are IDENTICAL.

When an arrangement includes identical elements, we must divide by the number of ways each set of identical elements can be arranged.

The reason:

When the identical elements swap positions, the arrangement doesn’t change, reducing the total number of unique arrangements.

Here, we must divide by the number of ways to arrange the 4 identical blue gumballs (4!) and the number of ways to arrange the 2 identical green gumballs (2!):

$\frac{7!}{4!2! } = 105$

Ans: B

Since, one child cannot have more than two brands of chocolates, therefore, one child can get a maximum of six chocolates.

When there are four winners, the winners can have the chocolates in the following possible way

Case I

One of them gets six chocolates and the other three get one each.

The one who gets six chocolates can be selected in 4C1 ways and he will get any two brands out of three in 3C2 ways.

The three chocolates of the same brand will be distributed among the other three in 3! ways.

Hence, the total number of ways = 4 × 3 × 6 = 72

Case II

One of them gets 5 chocolates, one of them gets 2 chocolates and the other two get 1 each.

Kid with 5 chocolates:

The one who gets five chocolates can be selected in 4C1 ways and he will get any two brands out of three in 3C2 ways.

After choosing the two brands (say A & B), he can get all three chocolates of one brand in 2 ways (i.e. either A or B). The other two chocolates can be chosen from the second brand in 3 ways. Thus, there are 4 × 3 × 2 × 3 = 36 ways

Kid with 2 chocolates:

This person can be chosen in 3 ways. Now, he can either get two chocolates from the same brand or from two different brands.

Thus, there are 6 × 3 = 18 ways

Kids with 1 chocolate each

The remaining chocolates can be distributed between the two remaining winners in 2 ways.

Therefore, total number of ways = 36 × 18 × 2 = 1296 ways.

Case III

Similarly for four chocolates the following combinations are possible:

a) 4 3 1 1

The number of ways = 4 × 3 × [{6 × 3(1 + ( 6 + 3))} + {1 × 9 × 3 (1 + 6)}] × 2 = 8856

b) 4 2 2 1

The number of ways = 4 × 3 × [{6 × 3(2 × 6 × 3 + (3 + 9))} + {9 × (9 + 9)}] = 12312

Since, the best of them can’t have less than 4 chocolates the above are the only cases applicable.

Hence, the total number of ways = 72 + 1296 + (8856 + 12312) = 22536.

Ans: B

The digit 6 is used once in 100~109 (only when 106).

Similarly the digit 6 is used once in 110~119(only when 116),

….

in 150~159(only when 156),

in 170~179(only when 176),

in 190~199(only when 196).

On the other hand in 160~169 the digit 6 is used at unit digit(166) and at tenth digit(16□).

So the digit 6 is used 11 times(=1 (unit digit)+10 (tenth digit)) in 160~169.

That means the digit 6 is used 20 in 100~199.

Similarly in 200~299 : 20 times

300~399 : 20 times

….

500~599 : 20 times

700~799 : 20 times

….

800~899 : 20 times

900~999 : 20 times

On the other hand in 600~699 we should consider hundredth digit so 6 is 100 times more used.

That means the digit 6 is used 120 in 600~699.

So total number of digit 6 is 20*10 + 100= 300.

Ans: C

Three digit number can have only following 3 patterns:

A. all digits are distinct;

B. two digits are alike and third is different;

C. all three digits are alike.

We need to calculate B. B=Total – A – C

Total numbers from 700 to 999 = 299 (3-digit numbers greater than 700);

A. all digits are distinct = 3*9*8=216 (first digit can have only three values 7, 8, or 9);

C. all three are alike = 3 (777, 888, 999).

So, 299-216-3=80.