# PE 2 - Permutation & Combination | Modern Math - Permutation & Combination

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**PE 2 - Permutation & Combination | Modern Math - Permutation & Combination**

There are seven markers of different colours, each with a body and a cap. Colour of the cap and the body of the marker is the same. What is the total number of ways in which the caps can be put on the body of the markers such that colour of the cap and the body for at least 1 marker is same?

Answer: 3186

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**Explanation** :

The total number of ways in which 7 caps can be put on 7 different markers is 7! = 5040.

Number of ways of putting caps on markers such that cap and body colour does not match for any marker = $7!\times \left[1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!}+\frac{1}{6!}-\frac{1}{7!}\right]$ = 2520 – 840 + 210 – 42 + 7 – 1 = 1854,

∴ Number of ways in which the caps can be put on the body of the markers such that colour of the cap and the body for at least 1 marker is the same = 5040 – 1854 = 3186.

Hence, 3186.

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**PE 2 - Permutation & Combination | Modern Math - Permutation & Combination**

How many 5-digit numbers have all its digits even?

- (a)
325

- (b)
450

- (c)
625

- (d)
1250

- (e)
2500

Answer: Option E

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**Explanation** :

We have to form a 5-digit number using digits 0, 2, 4, 6 and 8.

The first digit can be filled in 4 ways i.e., 2 or 4 or 6 or 8

The remaining digits can be filled in 5 ways i.e., 0 or 2 or 4 or 6 or 8

∴ Total number of 5-digit numbers with all its digits as even = 4 × 5 × 5 × 5 × 5 = 2500

Hence, option (e).

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**PE 2 - Permutation & Combination | Modern Math - Permutation & Combination**

In how many ways can 4 books on quant, 3 books on reasoning and 2 on verbal, all by different publishers, be arranged on a shelf?

- (a)
336880

- (b)
462

- (c)
40620

- (d)
80640

- (e)
362880

Answer: Option A

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**Explanation** :

As all the books are from different publishers, they are all different.

∴ Required number of ways to arrange the books = Total number of ways to arrange (4 + 3 + 2 =) 9 different books on a shelf = 9! = 362880

Hence, option (a).

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**PE 2 - Permutation & Combination | Modern Math - Permutation & Combination**

There are 6 girls and 4 boys to be seated in a circle. In how many ways can they be seated if four particular girls always want to sit together?

- (a)
7!

- (b)
7! × 4!

- (c)
9! × 4!

- (d)
6! × 4!

- (e)
9!

Answer: Option D

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**Explanation** :

Considering the 4 girls who want to sit together to be a single person, there will be 7 persons now. They can be arranged in 6! ways.

The 4 girls can be arranged among themselves in 4! ways.

∴ Required number of ways = 6! × 4!

Hence, option (d).

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**PE 2 - Permutation & Combination | Modern Math - Permutation & Combination**

In how many ways 6 friends can stay in four different rooms in a hotel such that all rooms are occupied.

Answer: 1560

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**Explanation** :

Since all rooms must be occupied, this can happen in two ways. 3, 1, 1, 1 or 2, 2, 1, 1.

**Case 1**: 3, 1, 1, 1 i.e., out 6 friends 3 will stay together and other 3 will stay individually in different rooms.

The 3 friends who stay together can se selected in ^{6}C_{3} = 20 ways.

Now, we have 4 different groups of friends (i.e., 3, 1, 1, 1) who need to stay in 4 different rooms. This can be done in 4! = 24 ways.

∴ Total number of ways = 20 × 24 = 480.

**Case 2**: 2, 2, 1, 1 i.e., out 6 friends 2 will stay together, another pair of 2 will stary in other rooms and 2 remaining friends will stay individually in different rooms.

The 2 groups of 2 friends each who stay together can se selected in ^{6}C_{2} × ^{4}C_{2} ÷ 2 = 45 ways.

Now, we have 4 different groups of friends (i.e., 2, 2, 1, 1) who need to stay in 4 different rooms. This can be done in 4! = 24 ways.

∴ Total number of ways = 45 × 24 = 1080.

⇒ Total number of ways 6 friends can stay in 4 different rooms = 480 + 1080 = 1560.

Hence, 1560.

Workspace:

**PE 2 - Permutation & Combination | Modern Math - Permutation & Combination**

The term independent of x in the expansion of ${\left[{x}^{2}-\frac{1}{x}\right]}^{12}$ is

- (a)
^{12}C_{6} - (b)
^{12}C_{3} - (c)
^{12}C_{4} - (d)
^{12}C_{5}

Answer: Option C

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**Explanation** :

We know:

(a + b)n = ^{n}C_{0} × a^{0} × b^{n} + ^{n}C_{1} × a^{1} × b^{n-1} + ^{n}C_{2} × a^{2} × b^{n-2} + … + ^{n}C_{n} × a^{n} × b^{0}

Let the term independent of x in ${\left[{x}^{2}-\frac{1}{x}\right]}^{12}$ is ^{n}C_{k} × (x^{2})^{k} × ${\left(\frac{1}{x}\right)}^{n-k}$

= ^{n}C_{k} × x^{2k} × x^{k-n}

= ^{n}C_{k} × x^{3k–n}

Now for this term to be independent of x, 3k – n = 0

⇒ k = n/3 = 12/3 = 4

∴ The term becomes ^{12}C_{4}.

Hence, option (c).

Workspace:

**PE 2 - Permutation & Combination | Modern Math - Permutation & Combination**

In how many ways can 4 persons, out of 5 couples, be selected, such that there should be no pair (couple)?

- (a)
48

- (b)
72

- (c)
80

- (d)
90

Answer: Option C

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**Explanation** :

We need to select 4 people out of 5 couples such that no couple is selected i.e., at max only one person will be selected from a couple.

First, we select 4 couples and then we will pick one from each of these 4 couples.

Number of ways of selecting 4 couples = ^{5}C_{4} = 5 ways.

From each of these 4 couples either the female or the male could be selected i.e., 2 ways of selection.

∴ Total number of ways of selecting 4 people = 5 × (2)^{4} = 80.

Hence, option (c).

Workspace:

**Answer the next 3 questions based on the information given below.**

India and England are playing a one-day cricket tournament, which contains 5 matches. There are 2 points for a win, 1 point for a draw and 0 point for loss. The team which scores more points wins the tournament.

**PE 2 - Permutation & Combination | Modern Math - Permutation & Combination**

If there is no draw, in how many different ways (i.e. by winning in different matches) can India win the tournament?

- (a)
16

- (b)
8

- (c)
10

- (d)
12

Answer: Option A

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**Explanation** :

**Case 1**: Indian wins all 5 matches

This can happen in only 1 way.

**Case 2**: Indian wins 4 matches and loses 1 match i.e., WWWWL

Number of ways = 5!/4! = 5 ways

**Case 3**: Indian wins 3 matches and loses 2 matches i.e., WWWLL

Number of ways = 5!/3!2! = 10 ways

∴ Total number of ways = 1 + 5 + 10 = 16 ways

Hence, option (a).

Workspace:

**PE 2 - Permutation & Combination | Modern Math - Permutation & Combination**

If exactly one match was drawn, in how many ways can India win the tournament?

- (a)
6

- (b)
10

- (c)
25

- (d)
30

Answer: Option C

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**Explanation** :

**Case 1**: Indian wins 4 matches and draws 1 match i.e., WWWWD

Number of ways = 5!/4! = 5 ways

**Case 2**: Indian wins 3 matches, loses 1 match and draws 1 match i.e., WWWLD

Number of ways = 5!/3! = 20 ways

∴ Total number of ways = 5 + 20 = 25 ways

Hence, option (c).

Workspace:

**PE 2 - Permutation & Combination | Modern Math - Permutation & Combination**

If any number of matches can be drawn, in how many ways can India win the tournament?

- (a)
100

- (b)
96

- (c)
64

- (d)
36

Answer: Option B

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**Explanation** :

**Case 1**: Indian wins all 5 matches i.e., WWWWW

This can happen in only 1 way.

**Case 2 (a)**: Indian wins 4 matches and loses 1 match i.e., WWWWL

Number of ways = 5!/4! = 5 ways

**Case 2 (b)**: Indian wins 4 matches and draws 1 match i.e., WWWWD

Number of ways = 5!/4! = 5 ways

**Case 3 (a)**: Indian wins 3 matches and loses 2 matches i.e., WWWLL

Number of ways = 5!/3!2! = 10 ways

**Case 3 (b)**: Indian wins 3 matches, loses 1 match and draws 1 match i.e., WWWLD

Number of ways = 5!/3! = 20 ways

**Case 3 (c)**: Indian wins 3 matches and draws 2 matches i.e., WWWDD

Number of ways = 5!/3!2! = 10 ways

**Case 4 (a)**: Indian wins 2 matches, loses 1 match and draws 2 matches i.e., WWLDD

Number of ways = 5!/2!2! = 30 ways

**Case 4 (b)**: Indian wins 2 matches, and draws 3 matches i.e., WWDDD

Number of ways = 5!/2!3! = 10 ways

**Case 5**: Indian wins 1 match, and draws 4 matches i.e., WDDDD

Number of ways = 5!/4! = 5 ways

∴ Total number of ways = 1 + 5 + 5 + 10 + 20 + 10 + 30 + 10 + 5 = 96 ways

Hence, option (b).

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