5 inspectors are sent to 3 stores, and each store must have at least one inspector, how many ways for there?

Answer 1

Call the stores A, B, C.

If you ignore the restriction of having at least one inspector per store, there are 3^5 = 243 ways to send the inspectors to the stores (each of the 5 inspectors has one of 3 choices).

Of these possibilities, we remove the 2^5=32 ways of leaving store A empty (each inspector has only the choice between B and C).

We also remove 31 ways of leaving store B empty (all except the one case where all inspectors go to store C, which has already been removed because it also leaves A empty).

And we remove the 30 remaining cases that leave C empty (except the cases where all go to A and all go to B).

Answer: 243 - 32 - 31 - 30 = 150.

Scythian was close but removed the three "all inspectors go to the same store" cases twice too many. It should be 243 - 96 + 3 (because they're already counted twice in the 96).

Reined would have got it right if he'd used 3! = 6 instead of 3! = 3 in the second part. 60/3! = 10, not 20.

Answer 2

You need to put 1 inspector in each store

you have 2 extras that you need to put somewhere

either you put it in the same store or you put it on different store
you get to possible combinations
2,2,1
3,1,1

First lets look at the combination 2,2,1
Distribute the inspector by store
For the first store, you have to choose 2 from 5 inspectors;
you have 5 choices for the first and 4 for the second: 5*4=20
but you have to divide by 2!
since having inspector 1 and 2 is the same thing than having inspector 2 and 1.
That give you 20/2=10
For the second store, You have to choose 2 from 3 inspectors;
you have 3 choices for the first and 2 for the second: 3*2=6
but you have to divide by 2! again
That give you 6/2=3
For the store 3 you only have 1 choice.
This gives you 10*3*1=30 choices

One store is going to have 1 inspector then you can say there is 3 choices for this store
That gives you 3*30= 90


Now the combination 3,1,1
Distribute the inspector by store
For the first store, you have to choose 3 from 5 inspectors;
you have 5 choices for the first and 4 for the second and 3 for the third: 5*4*3=60
but you have to divide by 3!
That give you 60/3=20
For the second store, You have to choose 1 from 2 inspectors: 2 choices;
For the store 3 you only have 1 choice.
This gives you 20*2*1=40 choices

One store is going to have 3 inspectors then you can say there is 3 choices for this store
That gives you 3*40= 120

Total 120+90=210

Answer 3

There are 3^5 = 243 ways to assign them, any number at any store. But there are (3)(2^5) ways to assign them that leaves 1 store without an inspection, and (3) ways to assign them that leaves 2 stores without an inspection. The answer therefore is:

243 - 96 - 3 = 144

Addendum: Benoit is right, I overlooked this. It's (3)(2^5) - 2 ways to assign them that leaves EXACTLY 1 store without an inspection, and so the answer is 243 - 90 - 3 = 150

Answer 4

60 ways if the 2 inspectors who are left do not take a store.

If they do its more, ill work it out