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

Hint: 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,
\[\begin{array}{l}
{a_1} = {a_2} = {a_3} = {a_4} = {a_5} = 3\\
m = 15
\end{array}\]
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.