A piggy bank contains 50 identical nickels, 50 identical dimes, and 50 identical quarters. How many ways are there to choose a collection of 20 coins?

you keep saying the coins are identical. does that mean the answer will be the same if there are 20 identical nickels, 20 identical dimes, and 20 identical quarters?

The way I understand your problem is : how many different ordered sums of 3 natural numbers (including zero) equal 20 ?

20 = 0 + 0 + 20 (1 possibility)

20 = 0 + 1 + 19 = 1 + 0 + 19 (2 possibilities)

20 = 0 + 2 + 18 = 1 + 1 + 18 = 2 + 0 + 18 (3 possibilities)
...
20 = 0 + 20 + 0 = 19 + 1 + 0 = ... = 20 + 0 + 0 (21 possibilities)

Answer is 1 + 2 + ... + 21 = 242

It has nothing to do with combinations, or with the given amount of available coins (50 of each), as soon as there are at least 20 of each.

Another question would be :
Suppose you pick up 20 coins at random, all coins among the 150 the bank holds having the same chance of being chosen.

... would be C(150,20).

How many ways are there to choose a collection of 20 coins including X nickels, Y dimes and Z quarters (X+Y+Z=20) ?
... would be C(50,X).C(50,Y).C(50,Z)

I took the problem a different way:
I think it should be done more like a password. A password with 20 letters, and the letters have to be chosen from the list = [n,d,q] (with replacement, from feanur's observation there is more than 20 of each kind of coin/letter).
Seems like 3^20 to me.

Wouldn't it be 50nPr20?

Lets do a few trivial cases:
use notation: nickel (n), dime (d), or quarter (q).
--------------------------
If you want to pull 0 coins: there is only 1 way to do that, you could write that is 3^0
--------------------------
If you want to pull 1 coin:
n
d
q
There are 3 way to do it, or 3^1
--------------------------
If you want to pull 2 coins:
n n
n d
n q

d n
d d
d q

q n
q d
q q

That is 9 ways, or 3^2.
--------------------------
If you want to pull 3 coins:

n n n
n n d
n n q
n d n
n d d
n d q
n q n
n q d
n q q

d n n
d n d
d n q
d d n
d d d
d d q
d q n
d q d
d q q

q n n
q n d
q n q
q d n
q d d
q d q
q q n
q q d
q q q

That is 27 ways or 3^3
--------------------------
My argument is if you want to pull 20 coins it is 3^20

Do you consider :
n d
d n
... as different picks ?

Anyway, I think the poster should elaborate a little bit.

Yes I did consider them as separate outcomes, and yes I'm not 100% certain that is what the problem is asking; but that would be my interpretation of what it asked.

The fact that we have 50 of each coin is irrelevant if we can only choose 20 coins. We don't need to worry about the number of coins in the jar unless we want the probability of a single outcome. We can just say that it is possible to have 20 of each coin in an outcome with none of the others because that is possible by the given parameters.


For 20 nickels, there is one combination
For 19 nickels, there are 2 combinations
For 18 nickels, there are 3 combinations
For 17 nickels, there are 4 combinations....
For 16 nickels, there are 5 combinations (16 4 0, 16 0 4, 16 3 1, 16 1 3, 16 2 2)
For 15 there are 6
For 14 there are 7
For 13 there are 8
For 12 there are 9 (12 8 0, 12 0 8, 12 7 1, 12 1 7, 12 6 2, 12 2 6, 12 5 3, 12 3 5, 12 4 4)....etc
For 11 there are 10
For 10 there are 11
For 9 there are 12
For 8 there are 13
For 7 there are 14
For 6 there are 15
For 5 there are 16
For 4 there are 17
For 3 there are 18
For 2 there are 19
For 1 there are 20
For 0 there are 21
sum them up gets 231

We don't need to do this again for the other 2 coin types as it would be a repeat in combinations.

This is combinations. If you care about permutations it is very simply 3^20 (3486784401).

This post was edited by Dontrunaway on Apr 1 2016 02:50pm

I'm not sure which is right, 231 or 230. Wolfram says 230 so I'm inclined to believe it.