+4 Find the number of ways that Magnus can give out 12 identical stickers to 4 of his friends. (Not everyone has to get a sticker.)
+527 Let's use casework!
We can consider different cases based on how many friends receive stickers:
No friends receive stickers: There is only 1 way to do this (give none to anyone).
One friend receives stickers: There are 4 ways to choose which friend gets stickers (out of the 4 friends), and then there is 1 way to distribute all 12 stickers to that one friend (since they are identical). So, there are 4 * 1 = 4 ways for this case.
Two friends receive stickers: There are 4 ways to choose which two friends get stickers, then we can distribute the stickers in (x12) ways, where x is the number of stickers given to the first friend (the remaining 12 - x stickers go to the second friend).
The number of ways to distribute the stickers like this is the same for any choice of which friend gets x stickers, so we sum over all possible values of x from 1 to 11. This gives us a total of:
4 * (11 + 10 + ... + 1) = 4 * \frac{11 \cdot 12}{2} = 264
Three friends receive stickers: Similar to the case of two friends, there are 4 ways to choose which three friends get stickers, and then we can distribute the stickers in (x12) ways for various values of x (number of stickers to the first friend). Summing over all possibilities for x gives us:
4 * (11 + 10 + ... + 2) = 4 * \frac{11 \cdot 10}{2} = 220
Four friends receive stickers: There are 4 ways to choose which four friends get stickers, and then 1 way to distribute all 12 identical stickers among them. So, there are 4 * 1 = 4 ways for this case.
Adding the number of ways for each case, we get a total of:
1 + 4 + 264 + 220 + 4 = 493
Therefore, there are 493 ways for Magnus to distribute his stickers.
+118677 Thanks Pythagorean.
I think I will try this one too.
I'll use a method called stars and bars.
Put the 12 stickers in a row (they are the stars)
Now we need to divide them into 4 groups. We can do this using 3 bars.
eg X X | X X X X | | X X X X X X represents the outcome of 2, 4, 0, 6
The question now becomes: How many positions can the 3 bars be placed in
The answer is 15C3 = 455 ways
Our answers are close Pythagoras but I think you have double counted some. ..
