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how many ways can you place 9 marbles in 3 hats so that a) each hat has at least 1 marble b) a hat may not have any marbles.

I'll take a shot...

You have 9 marbles total, and all marbles are exactly the same. You want at least one marble in each hat...so let's work it out.

Say you have Hat (A, B, C) If hat A gets 1 marble, you have (1, 1, 7) (1, 2, 6) (1, 3, 5) (1, 4, 4) (1, 5, 3) (1, 6, 2) (1, 7, 1) Total of 7

If hat B gets 2, you have (2, 1, 6) (2, 2, 5) (2, 3, 4) (2, 4, 3) (2, 5, 2) (2, 6, 1) Total of 6

If you see the pattern, you have... If A=1, Total = 7 ways If A=2, Total = 6 ways If A=3, Total = 5 ways ... If A=7, Total = 1 way You know that A cannot be more than 7 because other hats will not get any marbles. So number of marbles in A stops at 7. Now add up the total to get the answer for (a)... 7+6+5+4+3+2+1 = 28 ways

For (b), you have hat A with 0 marbles, do the same thing, you get (0, 1, 8) ... (0, 8, 1) Total of 8 Do the same if hat B has 0 and hat C has 0 marbles, you get a total of 8*3 = 24 ways

how many ways can you place 9 marbles in 3 hats so that a) each hat has at least 1 marble b) a hat may not have any marbles.

I'll take a shot...

You have 9 marbles total, and all marbles are exactly the same. You want at least one marble in each hat...so let's work it out.

Say you have Hat (A, B, C) If hat A gets 1 marble, you have (1, 1, 7) (1, 2, 6) (1, 3, 5) (1, 4, 4) (1, 5, 3) (1, 6, 2) (1, 7, 1) Total of 7

If hat B gets 2, you have (2, 1, 6) (2, 2, 5) (2, 3, 4) (2, 4, 3) (2, 5, 2) (2, 6, 1) Total of 6

If you see the pattern, you have... If A=1, Total = 7 ways If A=2, Total = 6 ways If A=3, Total = 5 ways ... If A=7, Total = 1 way You know that A cannot be more than 7 because other hats will not get any marbles. So number of marbles in A stops at 7. Now add up the total to get the answer for (a)... 7+6+5+4+3+2+1 = 28 ways

For (b), you have hat A with 0 marbles, do the same thing, you get (0, 1, 8) ... (0, 8, 1) Total of 8 Do the same if hat B has 0 and hat C has 0 marbles, you get a total of 8*3 = 24 ways

thanks for your response...in asking that question, i was hoping to find out whether there is a short-cut way to determine the number of ways you could split X objects amongst Y containers. I think I have seen a really elegant solution to such a problem on this forum however I cant seem to find it now...

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