Question

Conceptional question: You need to answer my confusion other than just solving the problem. Better follow the comment

Q1. 10 students are divided into 5 groups and each group will be assigned a project topic

(a) if the 5 groups will work on 5 different topics, find the number of ways of forming groups (why I can't use the multinomial coefficient to calculate Question 1???????)

answer: 10C2*8C2*6C2*4C2*2C2=113, 400

Example: A police department in a small city consists of 10 officers. If the department policy is
to have 5 of the officers patrolling the streets, 2 of the officers working full time at the
station, and 3 of the officers on reserve at the station, how many different divisions
of the 10 officers into the 3 groups are possible?

Why I can use the multinomial coefficient to calculate this one? answer: 10!/(5!2!3!)

Answer 1

The short answer to your query would be that you can use both the methods! Both the methods yield the same answer.

We know that if we want to divide (m + n + p) objects into 3 groups having m, n and p objects respectively, then the no. of ways to do this is = (m + n + p)!/(m! * n! * p!).

Here, in Q1, the no. of ways of forming groups = = 113400. (multinomial coefficient method) (since we are trying to make 5 groups containing 2 members each.

Also, no.of ways = 10C2 * 8C2 * 6C2 * 4C2 * 2C2=113400 (method of combinations).
In the above method, we first select 2 members from 10 members in 10C2 ways, then select 2 members from remaining 8 members in 8C2 ways, then select 2 members from remaining 6 members in 6C2 ways, then select 2 members from remaining 4 members in 4C2 ways and lastly select 2 members from remaining 2 members in 2C2 ways.

In the same way, in the example, no. of ways = 10!/(5! * 2! * 3!) = 10C5 * 5C2 * 3C3 = 2520.