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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.)

May 27, 2024

#1
+415
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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.

May 27, 2024
#2
+118623
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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. ..

May 28, 2024