4. On friday morning a pro shop has 14 cans of tennis balls. if they were all sold by sunday night, how many different ways could they have been sold on Friday Sat. ans Sunday?

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- Feb 11th 2009, 04:18 PMmsfynestriniHow many different ways
4. On friday morning a pro shop has 14 cans of tennis balls. if they were all sold by sunday night, how many different ways could they have been sold on Friday Sat. ans Sunday?

- Feb 12th 2009, 07:43 PMmeymathis
I will assume the cans are indistinguishable. Let's take it day by day. Let F be the number sold on Friday, and SA be the number sold on Saturday, and SU the number sold on Sunday.

Friday: 0 through 14 cans could be sold

Saturday: if Friday was 0, then there are 15 different possibilities (0-14)

if Friday was 1, then there are 14 different possibilities (0-13)

etc.

Sunday: SU = 14 - S - F. in other words, given what happened on Friday and Saturday, what happened on Sunday is already determined (since you know 14 total were sold).

So now we can add up the cases:

if F = 0: 15 cases

if F = 1: 14 cases

if F = 2: 13 cases

.

.

.

if F = 13: 2 case

if F = 14: 1 case

So the total number of cases is the sum of 1 through 15: (15*16/2) = 120. - Feb 12th 2009, 09:03 PMmeymathis
By the way. Like your other problem, it can be solved:

To distribute N items to M bins: $\displaystyle \mathbf{C}^{N+M-1}_{N} = {16 \choose 14}=\frac{16!}{14!\cdot 2!}=120 $