Question

6 letters are posted in three letter boxes. The number of ways of posting the letters when no letter box remains empty.
$
  {\text{A}}{\text{. 270}} \\
  {\text{B}}{\text{. 540}} \\
  {\text{C}}{\text{. 537}} \\
  {\text{D}}{\text{. None of these}} \\
$

Answer 1

Hint – To find the number of ways, we compute the number of ways in which we can arrange the 6 letters in 3 boxes and find their number of possibilities. We find the sum of all the possibilities of these cases to find the answer.

Complete Step-by-Step solution:
Given Data, 6 letters are posted in three letter boxes.
Let the boxes be A, B and C respectively.
The number of ways in which 6 letters can be added into 3 boxes is 3 + 2 + 1 or 2 + 2 + 2.
Case -1: (3 + 2 + 1)
We select one post box and put 3 of the 6 letters in that post box, then we select another postbox from the remaining and put 2 letters of the remaining 3 letters into it.
Number of ways = $\left( {{}^6{{\text{C}}_3} \times {}^3{{\text{C}}_1}} \right) \times \left( {{}^3{{\text{C}}_2} \times {}^2{{\text{C}}_1}} \right) = {\text{ 360}}$
Case -2: (2 + 2 + 2)
We select one post box and put 2 of the 6 letters in that post box, then we select another postbox from the remaining and put 2 letters of the remaining 4 letters into it.
Number of ways = $\left( {{}^6{{\text{C}}_2} \times {}^3{{\text{C}}_1}} \right) \times \left( {{}^4{{\text{C}}_2} \times {}^2{{\text{C}}_1}} \right) = {\text{ 180}}$

Hence the total number of ways = 360 + 180 = 540 ways, Option B is the correct answer.

Note – In order to solve such types of problems the key is to identify that the arrangement is a combination (order doesn’t matter). We have to carefully figure out the number of possibilities each of the arrangements in a box takes. To calculate the number of favorable outcomes, we need to use the combination formula${}^{\text{n}}{{\text{C}}_{\text{r}}} = \dfrac{{{\text{n!}}}}{{\left( {{\text{n - r}}} \right)!{\text{ r!}}}},{\text{ where n! means n}} \times {\text{(n - 1)}}... \times {\text{(n - (n - 1))}}$.