Question

In how many ways can 9 different books be distributed among three students if each receives at least 2 books.
A). 11085
B). 11805
C). 11508
D). None of these

Answer 1

Hint: To solve this question, we should have some knowledge of combinations. The combination can be defined as the number of ways of choosing some out of a 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 ${\displaystyle \frac{n!}{r!\left(n-r\right)!}}$.

Complete step-by-step solution:
In this question, we have to find the number of ways of distributing 9 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 5 books to anyone 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 3 books to one and 4 books to others. The third way is to distribute 3-3 books to all 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. The combination is a method of finding the number of ways of choosing some out of the whole, irrespective of their order. We can represent the formula of combination as ${}^n{C}_r={\displaystyle \frac{n!}{r!\left(n-r\right)!}}$.
Now, we will consider case 1, which is giving 4 books to anyone, and from the remaining 5 books, 2 - 2 books are given to both.
Case 1: Let us consider the first student receives 2 books and the second student receives 2 books and the third student receives 5 books. So, we can write,
$\Rightarrow$ Number of ways of distributing books $={}^9{C}_2\times {}^7{C}_2\times {}^5{C}_5$
We know that there are 6 possible cases of such type of distribution because there are 3 students. So, we can write,
$\Rightarrow$ Number of ways of distributing books $={}^9{C}_2\times {}^7{C}_2\times {}^5{C}_5\times 3!$
Now we will simplify each term using the formula,
${}^n{C}_r={\displaystyle \frac{n!}{r!\left(n-r\right)!}}$
$\Rightarrow$ Number of ways of distributing books $={\displaystyle \frac{9!}{2!7!}}\times {\displaystyle \frac{7!}{2!5!}}\times {\displaystyle \frac{5!}{5!0!}}\times 3!$
Expand the factorial,
$\Rightarrow$ Number of ways of distributing books $={\displaystyle \frac{9\times 8\times 7!}{2\times 1\times 7!}}\times {\displaystyle \frac{7\times 6\times 5!}{2\times 1\times 5!}}\times {\displaystyle \frac{5!}{5!0!}}\times 3!$
Simplify the terms,
$\Rightarrow$ Number of ways of distributing books $=36\times 21\times 1\times 3\times 2$
Multiply the terms,
$\Rightarrow$ Number of ways of distributing books $=4536$
Now, we will consider the second case, which is giving 2 books to any one of them and from the remaining 7 books, giving 3 books to the second and 4 to the third.
Case 2: Let us consider the first student receives 2 books and the second student receives 3 books and the third student receives 4 books. So, we can write,
$\Rightarrow$ Number of ways of distributing books $={}^9{C}_2\times {}^7{C}_3\times {}^4{C}_4$
We know that there are 6 possible cases of such type of distribution because there are 3 students. So, we can write,
$\Rightarrow$ Number of ways of distributing books $={}^9{C}_2\times {}^7{C}_3\times {}^4{C}_4\times 3!$
Now we will simplify each term using the formula,
${}^n{C}_r={\displaystyle \frac{n!}{r!\left(n-r\right)!}}$
$\Rightarrow$ Number of ways of distributing books $={\displaystyle \frac{9!}{2!7!}}\times {\displaystyle \frac{7!}{3!4!}}\times {\displaystyle \frac{4!}{4!0!}}\times 3!$
Expand the factorial,
$\Rightarrow$ Number of ways of distributing books $={\displaystyle \frac{9\times 8\times 7!}{2\times 1\times 7!}}\times {\displaystyle \frac{7\times 6\times 5\times 4!}{3\times 2\times 1\times 4!}}\times {\displaystyle \frac{4!}{4!0!}}\times 3!$
Simplify the terms,
$\Rightarrow$ Number of ways of distributing books $=36\times 35\times 1\times 3\times 2$
Multiply the terms,
$\Rightarrow$ Number of ways of distributing books $=7560$
Now, we will consider the third case, which is giving 3 – 3 books to each of them.
Case 3: Let us consider all three students receive 3 books each of them. So, we can write,
$\Rightarrow$ Number of ways of distributing books $={}^9{C}_3\times {}^6{C}_3\times {}^3{C}_3$
Now we will simplify each term using the formula,
${}^n{C}_r={\displaystyle \frac{n!}{r!\left(n-r\right)!}}$
We know that there are 6 possible cases of such type of distribution because there are 3 students. So, we can write,
$\Rightarrow$ Number of ways of distributing books $$={\displaystyle \frac{9!}{3!6!}}\times {\displaystyle \frac{6!}{3!3!}}\times {\displaystyle \frac{3!}{3!0!}}\times 3!$$
Expand the factorial,
$\Rightarrow$ Number of ways of distributing books $={\displaystyle \frac{9\times 8\times 7\times 6!}{3\times 2\times 1\times 6!}}\times {\displaystyle \frac{6\times 5\times 4\times 3!}{3\times 2\times 1\times 3!}}\times {\displaystyle \frac{3!}{3!0!}}\times 3!$
Simplify the terms,
$\Rightarrow$ Number of ways of distributing books $=84\times 20\times 1\times 6$
Multiply the terms,
$\Rightarrow$ Number of ways of distributing books $=10080$
Now, we have to find the total number of ways from the three cases. So, we will get,
$\Rightarrow$ Total number of ways of distributing the books $=4536+7560+10080$
Add the terms,
$\Rightarrow$ Total number of ways of distributing the books $=22176$
Hence, option (D) is the correct answer.

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