# CRE 7 - Letters and Letter Boxes | Modern Math - Permutation & Combination

**Answer the next 4 questions based on the information given below:**

There are 4 letters and 3 letter boxes. In how many ways can letters be posted in these letter boxes if

**CRE 7 - Letters and Letter Boxes | Modern Math - Permutation & Combination**

All 4 letters are distinct, and 3 letter boxes are also distinct.

Answer: 4

**Explanation** :

Both letters and letter boxes are distinct here.

Each letter can be sent to any of the 3 letter boxes, i.e., in 3 ways.

∴ 4 distinct letters can be posted in 3^{4} = 81 ways.

Hence, 81.

Workspace:

**CRE 7 - Letters and Letter Boxes | Modern Math - Permutation & Combination**

All 4 letters are similar, and 3 letter boxes are also similar.

Answer: 4

**Explanation** :

Both letters and letter boxes are similar here.

Now, we will have to just do the grouping of letters.

LB LB LB

**Case 1** 4 0 0

**Case 2** 3 1 0

**Case 3** 2 1 1

**Case 4** 2 2 0

It does not matter which letter box has how many letter since letter boxes are all similar.

∴ There are 4 ways of posting 4 similar letters in 3 similar letter boxes.

Hence, 4.

Workspace:

**CRE 7 - Letters and Letter Boxes | Modern Math - Permutation & Combination**

All 4 letters are similar, but the 3 letter boxes are distinct.

Answer: 15

**Explanation** :

Here letters are similar, but the letter boxes are distinct.

Let the three distinct letter boxes get x, y and z letters.

∴ x + y + z = 4.

Note: This is same as distributing n identical item to r different people.

∴ The required answer = ^{4+3-1}C_{3-1} = ^{6}C_{2} = 15 ways.

Hence, 15.

Workspace:

**CRE 7 - Letters and Letter Boxes | Modern Math - Permutation & Combination**

All 4 letters are distinct, but the 3 letter boxes are similar.

Answer: 14

**Explanation** :

Here letters are distinct, but the letter boxes are similar.

Now, we first need to do the grouping of letters i.e.,

LB LB LB

Case 1 4 0 0

Case 2 3 1 0

Case 3 2 1 1

Case 4 2 2 0

**Case 1:** Since all 4 letters go in a single letter box, there is only 1 way of doing this.

**Case 2: **3 letters go in one of the letter boxes and 1 letter goes in another letter box.

Number of ways of selecting 3 letters out of 4 = ^{4}C_{3} = 4 ways. We send these 3 letters in any of the boxes (they are all similar) in 1 way. The 4^{th} letter can be sent in any of the remaining two letter boxes (similar) in 1 way.

∴ Total number of ways = 4 × 1 × 1 = 4 ways.

**Case 3: **2 letters go in one of the letter boxes and 2 other go in different letter boxes.

Number of ways of selecting 2 letters out of 4 = ^{4}C_{2} = 6 ways. We send these 2 letters in any of the boxes (they are all similar) in 1 way. The 3^{rd} and the 4^{th} letter each can be sent in the remaining two letter boxes (similar) in 1 way.

∴ Total number of ways = 6 × 1 × 1 = 6 ways.

**Case 4:** 2 letters go in on of the letter boxes and 2 others go in another letter box.

Number of ways of making 2 groups of 2 letters each = 3.

L_{1}L_{2} & L_{3}L_{4} or L_{1}L_{3} & L_{2}L_{4} or L_{1}L_{4} & L_{2}L_{3}.

∴ The required answer = 1 + 4 + 6 + 3 = 14 ways.

Hence, 14.

Workspace:

**CRE 7 - Letters and Letter Boxes | Modern Math - Permutation & Combination**

In how many ways can 12 distinct shirts be distributed among 3 people such that each person may / may not get a shirt?

- (a)
1548

- (b)
36

- (c)
12

^{3} - (d)
3

^{12}

Answer: Option D

**Explanation** :

Both shirts and people are distinct here.

First shirt can be given to any of the 3 persons i.e., in 3 ways.

Similarly, all other shirts can also be given to any of the 3 persons i.e., in 3 ways each.

Hence, total number of ways of distributing 12 distinct shirts among 3 people = 3^{12}.

Note: No. of ways of distributing distinct items to n distinct people = n^{m}.

Hence, option (d).

Workspace:

**CRE 7 - Letters and Letter Boxes | Modern Math - Permutation & Combination**

In how many ways can 12 identical shirts be distributed among 3 people such that each person may / may not receive a shirt?

Answer: 91

**Explanation** :

Here shirts are identical, and people are distinct.

Let the each of three persons receive x, y and z number of shirts from these 12 shirts.

∴ x + y + z = 12.

Note: This is same as distributing n identical item to r different people.

No. of ways of distributing n identical items to r distinct people = ^{n+r-1}C_{r-1}.

⇒ Number of ways 12 identical shirts can be given to 3 people such that each may / may not receive a shirt = ^{12+3-1}C_{3-1} = ^{14}C_{2} = 91 ways.

Hence, 91.

Workspace:

**CRE 7 - Letters and Letter Boxes | Modern Math - Permutation & Combination**

In how many ways can 12 identical shirts be distributed among 3 people such that each person gets at least 1 chocolate?

Answer: 55

**Explanation** :

Here shirts are identical, and people are distinct.

We need to give at least 1 shirt to all 3. Let us give 1 shirt to each of the three, hence we will have 9 shirts left. Now there is not restriction on distribution these 9 shirts.

Let the each of three persons receive x, y and z number of shirts from these 9 shirts.

∴ x + y + z = 9.

Note: This is same as distributing n identical item to r different people.

No. of ways of distributing n identical items to r distinct people = ^{n+r-1}C_{r-1}.

⇒ Number of ways 9 identical shirts can be given to 3 people such that each may / may not receive a shirt = ^{9+3-1}C_{3-1} = ^{11}C_{2} = 55 ways.

Hence, 55.

Workspace:

**CRE 7 - Letters and Letter Boxes | Modern Math - Permutation & Combination**

In how many ways can at most 12 identical shirts be distributed among 3 people such that each person may / may not get a chocolate?

Answer: 455

**Explanation** :

Here shirts are identical, and people are distinct.

Let the each of three persons receive x, y and z number of shirts.

The total number of shirts received is less than or equal to 12. Let’s say total number of shirts received is (12 – w).

∴ x + y + z ≤ 12 - w.

where, w ≥ 0.

⇒ x + y + z + w = 12.

Note: This is same as distributing n identical item to r different people.

No. of ways of distributing n identical items to r distinct people = n+r-1Cr-1.

∴ Required answer = ^{12+4-1}C_{4-1} = ^{15}C_{3} = 455.

Hence, 455.

Workspace:

**CRE 7 - Letters and Letter Boxes | Modern Math - Permutation & Combination**

In how many ways can 12 distinct shirts be distributed among 3 people such that each person receives at least 1 shirt?

- (a)
3

^{12}- 3069 - (b)
3

^{12} - (c)
3

^{12}- 3071 - (d)
None of these

Answer: Option A

**Explanation** :

Here shirts are similar, and people are distinct.

Total number of ways = 3^{12}

This includes those cases when 1 or 2 people don’t receive any shirt.

**Case 1:** 2 people don’t receive any thing i.e., 1 receives all the 12 shirts.

Number of ways of selecting 1 person out of 3 = ^{3}C_{2} = 3 ways.

Now number of ways of giving 12 shirts to this person = 1.

∴ Total number of ways in which only 1 person receives all the 3 shirts = 3 × 1 = 3 ways.

**Case 2:** 2 people receive at least 1 shirt, and 1 person doesn’t get any shirt.

Number of ways of selecting 1 person out of 3 = ^{3}C_{2} = 3 ways.

Now number of ways of giving 12 shirts to 2 other persons = 2^{12} - 2.

[2^{12} is total number of ways of distribution 12 shirts to 2 people. There will be 2 cases here where 1 person receive all 12 shirts.]

∴ Total number of ways in which only 2 persons receives all the 3 shirts = 3 × (2^{12} - 2) ways.

∴ Total number of ways in which every person receives at least 1 shirt = 3^{12} – 3 – 3(2^{12} - 2) = 3^{12} – 3 × 2^{12} + 3 = 3^{12} – 3069.

Hence, option (a).

Workspace:

## Feedback

Help us build a Free and Comprehensive Preparation portal for various competitive exams by providing us your valuable feedback about Apti4All and how it can be improved.