What is the solution to the problem in combinatorics?

Question

Mathematics

Ostap Bender has 100 elephants in a bag, of which k are multi-colored, and he distributes them to m children, one elephant in each hand. k < m < 100
In how many ways can the distribution of elephants take place?

Answer 1

Ostap Bender has 100 elephants in a bag, of which k are multi-colored, and he distributes them to m children, one elephant in each hand ( k < m < 100). In how many ways can the distribution of elephants take place?

The first elephant can be chosen in 100 ways and given away min ways.
Second choose 99 and give m – 1way.

N = 100 * m * 99 * (m-1) * ... * (100 - m) * 1
N = 100! / m! * m! = 100!

Total, the answer is 100!(100 factorial)
But it seems that I didn’t understand something in the condition of the problem - why are there different colors of some elephants?

Answer 2

We have k colorful elephants and 100-k identical elephants. Children are different, elephants of the same color are the same.
Head-on solution:
There are 2 variants of the problem:
1) m <= 100-k (there are enough identical elephants for all children)
2) m > 100-k (there are not enough identical elephants for all)
In the first case, you need to understand how many different ways you can distribute Since there are enough identical elephants for everyone in the first case, the number of ways can be as follows:
The number of ways to draw k_i colorful elephants:
C_{k}^{k_i}
The number of ways to get k_i colorful elephants out of the bag and distribute them to children:
C_{ k}^{k_i} A_{m}^{k_i}
There can be from 0 to k different-colored elephants pulled out of the bag, which means that in the end we get:
\sum_{i=0}^{i=k} C_{k}^{i} A_{m}^{i}
In the second case, in accordance with Derichlet's rule, at least (m + k - 100) colored bishops will be chosen, the rest is the same:
\sum_{i=m+k-100}^{i=k} C_{k}^{i} A_{m}^{i}
Total, in the general case:
\sum_{i=max( m+k-100, 0)}^{i=k} C_{k}^{i} A_{m}^{i}