Question

) We throw 9 identical balls into 7 bins. How many different ways are there to distribute these 9 balls among the 7 bins such that no bin is empty? Assume the bins are distinguishable (e.g., numbered 1 through 7).

Answer 1

Answer:

28 ways

Step-by-step explanation:

After placing 1 ball in each of the seven bins, there are two balls left.

If we place both balls in a single bin, there are 7 different ways to place the balls (place both on bins 1 through 7).

If we place each of the remaining balls in a different bin, the number of ways to place the balls is:

$n_2=\frac{7!}{(7-2)!2!}=7*3=21$

The total number of ways to distribute those balls is 21 + 7 = 28 ways.