Forming a band- permutations John, Jim, Jay, and Jack have formed a band consisting of 4 instruments. If each of the boys can play all 4 instruments, how many different arrangements are possible? What if John and Jim can play all 4 instruments, but Jay and Jack can each play only piano and drums?
If each of them can play all instruments, the number of arrangements will be 24 ways.
I don't know how to proceed for the second part
 A: The first part is correct! $\to$ There are $4$ distinct instruments to distribute to $4$ distinct people. And there are $4! = 24$ distinct ways of doing this.

For the second part. $\to$ Jay and Jack can only play $2$ instruments, therefore we have $2$ instruments for $2$ people, and the distribution for these guys can be made in $2! = 2$ distinct ways.
Then, note that all that is left are $2$ distinct instruments for $2$ people. So there is an additional $2$ distinct ways of doing this distribution for these other guys.
By the product rule, in total, we can do $4$ distinct distribution in this condition.
A: I see you've correctly answered the first part, i.e. when all four members play all four instruments, then the possible arrangements are 4! = 24.
For the second part we need to understand that as two instruments are only to be played by Jay and Jack, John and Jim also limit their choices to 2 instruments because although they can play all four instruments, piano and drums are always to be played by Jay and Jack. So for John and Jim, the possible number of arrangements becomes (4-2)! = 2 (because 2 instruments are already taken by Jay and Jack). For Jay and Jack as well, we see they only have the skill to play two instruments among themselves, so for them, the possible arrangement also becomes 2! = 2.
Hence, to answer the question, if Jay and Jack can only play 2 instruments, then the possible number of arrangements becomes 2! * 2! = 4. (by applying the product rule).
