Question

The number of ways to make 5 heaps of 3 books each from 15 different books is
A) $\dfrac{{15!}}{{\left( {5!} \right){{\left( {3!} \right)}^5}}}$
B) $\dfrac{{15!}}{{{{\left( {3!} \right)}^5}}}$
C) ${}^{15}{C_3}$
D) ${}^{15}{P_5}$

Answer 1

Here we will be using the concept of arrangement of distinct objects into equal groups.

Complete step by step answer:
5 heaps of 3 books each are to be made from 15 different books. We are to find in how many ways this can be done.

We know that the formula of dividing $m$ different things into groups of sizes ${a_1},{a_2},{a_3}...,{a_n}$ where ${a_1} + {a_2} + {a_3} + ... + {a_n} = m$is $\dfrac{{m!}}{{\left( {{a_1}!} \right)\left( {{a_2}!} \right)\left( {{a_3}!} \right)...\left( {{a_n}!} \right)}}$.
But this is applicable on if the groups are of unequal sizes.
In the given problem all the groups are of size 3, and there are 5 groups.
Hence,

{a_1} = {a_2} = {a_3} = {a_4} = {a_5} = 3\\\ m = 15 \end{array}$$ Again, the equal groups will be arranged amongst themselves, which is possible in $$5!$$ ways. Thus, we can say that the required number of ways in which the 15 different books can be divided into 5 groups of size 3 is $$\begin{array}{l} \dfrac{{15!}}{{\left( {3!} \right)\left( {3!} \right)\left( {3!} \right)\left( {3!} \right)\left( {3!} \right) \cdot \left( {5!} \right)}}\\\ = \dfrac{{15!}}{{{{\left( {3!} \right)}^5} \cdot \left( {5!} \right)}} \end{array}$$ **Note:** In this type of questions, it is to be always remembered that separate approaches need to be adopted for distinct and identical objects. Here, the books were distinct or different.