Question

number of ways of distributing 21 different objects in boxes of size 3 7 11

Answer 1

3527160

Answer 2

The problem asks for the number of ways to distribute 21 different objects into three distinguishable boxes of sizes 3, 7, and 11. Since the boxes have different specified sizes, they are distinguishable.

We can solve this problem by considering the process of selecting objects for each box sequentially.

Step 1: Select 3 objects out of the 21 different objects to be placed in the box of size 3.
The number of ways to do this is given by the combination formula:

$\binom{21}{3} = \frac{21!}{3!(21-3)!} = \frac{21!}{3!18!}$

Step 2: From the remaining $21 - 3 = 18$ objects, select 7 objects to be placed in the box of size 7.
The number of ways to do this is:

$\binom{18}{7} = \frac{18!}{7!(18-7)!} = \frac{18!}{7!11!}$

Step 3: The remaining $18 - 7 = 11$ objects must be placed in the box of size 11.
The number of ways to do this is:

$\binom{11}{11} = \frac{11!}{11!(11-11)!} = \frac{11!}{11!0!} = 1$

Since the choices for each box are independent, the total number of ways to distribute the objects is the product of the number of ways at each step:

Total ways = $\binom{21}{3} \times \binom{18}{7} \times \binom{11}{11}$
Total ways = $\frac{21!}{3!18!} \times \frac{18!}{7!11!} \times \frac{11!}{11!0!}$

We can see that the terms cancel out:

Total ways = $\frac{21!}{3! \times 7! \times 11!}$

This formula represents the number of ways to partition a set of 21 distinct objects into three ordered groups of sizes 3, 7, and 11. Since the boxes are distinguishable (by their sizes), assigning a specific group to a specific box of the corresponding size is a unique outcome. Thus, the number of ways to distribute the objects is indeed $\frac{21!}{3! 7! 11!}$.

The number of ways to distribute $n$ distinct objects into $k$ distinguishable boxes with capacities $n_1, n_2, \dots, n_k$ such that $n_1 + n_2 + \dots + n_k = n$ is given by the multinomial coefficient $\frac{n!}{n_1! n_2! \dots n_k!}$.
Here, $n=21$, $n_1=3$, $n_2=7$, $n_3=11$. The boxes are distinguishable by their sizes.
The number of ways is $\frac{21!}{3! 7! 11!} = \binom{21}{3} \binom{18}{7} \binom{11}{11} = 1330 \times 2652 \times 1 = 3527160$.