Aptitude Discussion

Q. |
In how many ways can $5$ different toys be packed in $3$ identical boxes such that no box is empty, if any of the boxes may hold all of the toys? |

✖ A. |
20 |

✖ B. |
30 |

✔ C. |
25 |

✖ D. |
600 |

✖ E. |
480 |

**Solution:**

Option(**C**) is correct

The toys are different; The boxes are identical

If none of the boxes is to remain empty, then we can pack the toys in one of the following ways

a. $2, 2, 1$

b. $3, 1, 1$

**Case A:**

Number of ways of achieving the first option $2 - 2 - 1$

Two toys out of the $5$ can be selected in ${^5C_2}$ ways. Another $2$ out of the remaining $3$ can be selected in ${^3C_2}$ ways and the last toy can be selected in ${^1C_1}$ way.

However, as the boxes are identical, the two different ways of selecting which box holds the first two toys and which one holds the second set of two toys will look the same. Hence, we need to divide the result by $2$.

Therefore, total number of ways of achieving the $2 - 2 - 1$ option is:

$= \dfrac{{^5C_2} × {^3C_2} }{2}$

$= \dfrac{10 × 3}{2}$

$= 15\text{ ways.}$

**Case B:**

Number of ways of achieving the second option $3 - 1 - 1$

Three toys out of the $5$ can be selected in${^5C_3}$ ways. As the boxes are identical, the remaining two toys can go into the two identical looking boxes in only one way.

Therefore, total number of ways of getting the $3 - 1 - 1$ option is ${^5C_3} = 10$ ways.

Total ways in which the $5$ toys can be packed in $3$ identical boxes

$= \text{number of ways of achieving Case A} + \text{number of ways of achieving Case B}$

$= 15 + 10$

$= 25\text{ ways.}$

**Amit**

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**Debasish Dey**

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Also if we go like 1 - 2 -2, then the calculation would be T = 5C1 x 4C2 x 2C2.

If boxes are different, we have to mark the boxes are 1 - 2 - 3 and there would T X 3P3 ways.

**Ashish**

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Can't the solution be 5*5*5/3 = 25 ??

how about 5c2 + 5c2 + 5c1 = 25?