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

Answer: 4

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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 3⁴ = 81 ways.

Hence, 81.

Answer: 4

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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.

Answer: 15

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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+3-1C_3-1 = ⁶C₂ = 15 ways.

Hence, 15.

Answer: 14

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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 = ⁴C₃ = 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 = ⁴C₂ = 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.

L₁L₂ & L₃L₄ or L₁L₃ & L₂L₄ or L₁L₄ & L₂L₃.

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

Hence, 14.

Answer: Option D

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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 = 3¹².

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

Hence, option (d).

Answer: 91

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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-1C_r-1.

⇒ Number of ways 12 identical shirts can be given to 3 people such that each may / may not receive a shirt = ^12+3-1C_3-1 = ¹⁴C₂ = 91 ways.

Hence, 91.

Answer: 55

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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-1C_r-1.

⇒ Number of ways 9 identical shirts can be given to 3 people such that each may / may not receive a shirt = ^9+3-1C_3-1 = ¹¹C₂ = 55 ways.

Hence, 55.

Answer: 455

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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+4-1C_4-1 = ¹⁵C₃ = 455.

Hence, 455.

Answer: Option A

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Explanation :

Here shirts are similar, and people are distinct.

Total number of ways = 3¹² 

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 = ³C₂ = 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 = ³C₂ = 3 ways.

Now number of ways of giving 12 shirts to 2 other persons = 2¹² - 2.

[2¹² 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 × (2¹² - 2) ways.

∴ Total number of ways in which every person receives at least 1 shirt = 3¹² – 3 – 3(2¹² - 2) = 3¹² – 3 × 2¹² + 3 = 3¹² – 3069.

Hence, option (a).