Question

In how many ways can 6 different books be distributed among 3 students such that none gets equal number of books and each gets atleast one book?

Answer 1

360

Answer 2

To solve this problem, we need to find the number of ways to distribute 6 different books among 3 students such that:

1. Each student receives at least one book.
2. None of the students receives an equal number of books.

Let the number of books received by the three students be $x_1, x_2, x_3$.

From the given conditions:

1. $x_1 + x_2 + x_3 = 6$
2. $x_1, x_2, x_3 \ge 1$ (each gets at least one book)
3. $x_1 \ne x_2$, $x_2 \ne x_3$, $x_3 \ne x_1$ (none gets an equal number of books)
4. $x_1, x_2, x_3$ are integers.

First, let's find all possible partitions of 6 into 3 distinct positive integer parts.

Possible partitions of 6 into 3 parts, where each part is at least 1:

• If we start with the smallest possible number for one student (1 book):

  • $1 + 1 + 4 = 6$ (Not allowed, as two students get 1 book, which is an equal number)
  • $1 + 2 + 3 = 6$ (Allowed, as 1, 2, and 3 are all different numbers)

• If we start with 2 books for one student (and ensure the other parts are also at least 1):

  • $2 + 2 + 2 = 6$ (Not allowed, as all students get 2 books, which is an equal number)

Thus, the only valid distribution of the number of books is (1, 2, 3). This means one student gets 1 book, another gets 2 books, and the third student gets 3 books.

Now, we need to distribute the 6 different books according to this partition. Since the students are distinct, we need to consider which student gets how many books.

Let the students be Student A, Student B, and Student C.

The distribution of the counts (1, 2, 3) can be assigned to the students in $3!$ ways (e.g., Student A gets 1, B gets 2, C gets 3; or Student A gets 2, B gets 1, C gets 3; and so on).

Consider a specific assignment, for example:

• Student A gets 1 book.
• Student B gets 2 books.
• Student C gets 3 books.

The number of ways to choose the books for this specific assignment is:

1. Student A chooses 1 book from 6: $\binom{6}{1}$ ways.
2. Student B chooses 2 books from the remaining $6-1=5$ books: $\binom{5}{2}$ ways.
3. Student C chooses 3 books from the remaining $5-2=3$ books: $\binom{3}{3}$ ways.

So, the number of ways for this specific ordered distribution (1 book for A, 2 for B, 3 for C) is:

$\binom{6}{1} \times \binom{5}{2} \times \binom{3}{3} = 6 \times \frac{5 \times 4}{2 \times 1} \times 1 = 6 \times 10 \times 1 = 60$ ways.

Alternatively, this can be calculated using the multinomial coefficient formula for distributing distinct items into distinct groups with specified sizes: $\frac{n!}{n_1! n_2! n_3!}$.

For the assignment (Student A gets 1, Student B gets 2, Student C gets 3), the number of ways is $\frac{6!}{1!2!3!} = \frac{720}{1 \times 2 \times 6} = \frac{720}{12} = 60$ ways.

Since the counts (1, 2, 3) are all distinct, there are $3!$ ways to assign these counts to the 3 distinct students.

The possible assignments of counts to (Student A, Student B, Student C) are permutations of (1, 2, 3):

• (1, 2, 3)
• (1, 3, 2)
• (2, 1, 3)
• (2, 3, 1)
• (3, 1, 2)
• (3, 2, 1)

There are $3! = 3 \times 2 \times 1 = 6$ such permutations.

For each of these permutations, the number of ways to distribute the actual books is 60.

Therefore, the total number of ways to distribute the books is $60 \times 6 = 360$.