Question

The number of different ways in which 8 different books can be distributed among 3 students, if each student receives at least 2 books is ?

Answer 1

Hint: To solve this question, we should have some knowledge of combinations. Combination can be defined as the number of ways of choosing some out of whole, irrespective of their order. Mathematically, it is represented by the term ${}^{n}{{C}_{r}}$, where r is the number of items to be chosen and n represents the total number of items in the event. ${}^{n}{{C}_{r}}$ is expanded as $\dfrac{n!}{r!\left( n-r \right)!}$.

Complete step-by-step solution -
In this question, we have to find the number of ways of distributing 8 different books among 3 students, if each student receives at least 2 books. To solve this question, we will consider all the possible cases. Among them, one is distributing 4 books to any one and from the remaining 4 books, 2 - 2 books to both. And the second way is distributing 2 books to any one of them and from the remaining books 3 - 3 books to each of them. Now, we have to find the number of ways of distributing books in each case. For that, we should have some knowledge of combination. Combination is a method of finding the number of ways of choosing some out of whole, irrespective of their order. We can represent the formula of combination as ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$.
Now, we will consider case 1, that is giving 4 books to any one and from the remaining 4 books, 2 - 2 books are given to both.
Case 1: Let us consider first student receives 2 books and second student receives 2 books and third student receives 4 books. So, we can write, number of ways of distributing books $={}^{8}{{C}_{2}}\times {}^{6}{{C}_{2}}\times {}^{4}{{C}_{4}}$. We know that there are 3 possible cases of such type of distribution, because there are 3 students. So, we can write, number of ways of distributing books according to case 1 as $={}^{8}{{C}_{2}}\times {}^{6}{{C}_{2}}\times {}^{4}{{C}_{4}}\times 3$. Now we will simplify each term using the formula, ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$. So, we get, $\begin{align}
  & \dfrac{8!}{2!\left( 8-2 \right)!}\times \dfrac{6!}{2!\left( 6-2 \right)!}\times \dfrac{4!}{4!\left( 4-4 \right)!}\times 3 \\
 & \Rightarrow \dfrac{8!}{2!6!}\times \dfrac{6!}{2!4!}\times \dfrac{4!}{4!0!}\times 3 \\
 & \Rightarrow \dfrac{8\times 7}{2\times 1}\times \dfrac{6\times 5}{2\times 1}\times 1\times 3 \\
 & \Rightarrow 1260 \\
\end{align}$
Now, we will consider the second case, that is giving 2 books to any one of them and from the remaining 6 books, giving 3 - 3 books to both of them.
Case 2: Let us consider the first student receives 2 books and the rest two students receive 3 books each. So, we can write, number of ways of distributing books $={}^{8}{{C}_{2}}\times {}^{6}{{C}_{3}}\times {}^{3}{{C}_{3}}$. Again, we can say that there are 3 possible cases of such types of distribution, because there are 3 students and, in each case, different students will get only 2 books. So, we can write, number of ways of distributing books according to case 2 as $={}^{8}{{C}_{2}}\times {}^{6}{{C}_{3}}\times {}^{3}{{C}_{3}}\times 3$. Now we will use the formula, ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ to simplify. So, we get,
$\begin{align}
  & \dfrac{8!}{2!\left( 8-2 \right)!}\times \dfrac{6!}{3!\left( 6-3 \right)!}\times \dfrac{3!}{3!\left( 3-3 \right)!}\times 3 \\
 & \Rightarrow \dfrac{8!}{2!6!}\times \dfrac{6!}{3!3!}\times \dfrac{3!}{3!0!}\times 3 \\
 & \Rightarrow \dfrac{8\times 7}{2\times 1}\times \dfrac{6\times 5\times 4}{3\times 2\times 1}\times 1\times 3 \\
 & \Rightarrow 1680 \\
\end{align}$
Now, we have to find the total number of ways from both the cases. So, we will get,
Total number of ways of distributing the books = Number of ways according to case 1 + number of ways according to case 2
= 1260 + 1680 = 2940
Hence, there are 2940 ways to distribute 8 different books among 3 students such that each student receives at least 2 books.

Note: The possible mistake one can make in this question is by considering all the books identical in a hurry, which will give the answer as 6, which is wrong and nowhere close to the answer. Also, there is a chance of forgetting to consider the case of giving 4 books to 1 student, which can also result in a wrong answer.