101 Dalmations!

Probability Level 4

Cruella De Vil has 15 dalmations captive. She wishes to cage them into 5 distinct cages, with each cage containing 3 puppies. If each dalmatian is distinct, and the number of ways Cruella can arrange the 15 dalmatians in each of the cages is n n , find the sum of the digits of n n .

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The answer is 30.

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Yan Yau Cheng by Yan Yau Cheng , 20, Hong Kong

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2 solutions

Eddie The Head
Apr 30, 2014

The answer will be ( 15 3 ) ∗ ( 12 3 ) ∗ ( 9 3 ) ∗ ( 6 3 ) ∗ ( 3 3 ) = 168168000 \dbinom{15}{3}*\dbinom{12}{3}*\dbinom{9}{3}*\dbinom{6}{3}*\dbinom{3}{3} = 168168000 .Hence the sum of the digits is 1 + 6 + 8 + 1 + 6 + 8 = 30 1+6+8+1+6+8 = \boxed{30} .

Note: \textbf{Note:} You should also have mentioned that the cages are distinct that is the 1 s t 1st 3 3 dalmatians going to cage 1 1 is different from them going to cage 2 2 (say).Had all the cages been similar we would have to divide the whole thing by 5 ! 5! .

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Yeah I assumed the cages were all similar, so I had an incorrect answer.

Matthew Mann - 7 years, 1 month ago

Easy and wonderful problem :)

Krishna Ar - 7 years, 1 month ago

Damn!!! I did not add the digits of 168168000.

Steven Zheng - 6 years, 11 months ago

Please do add that information to the question... Probably the reason the rating on this relatively simple problem is so high, everyone is getting it wrong because the question isn't clear. Just curious though, if the cages weren't distinct, what would be the answer? Would it just be divided by 5! ?

Nicolas Bryenton - 6 years, 11 months ago

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Thanks. I've updated the problem statement accordingly.

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Calvin Lin Staff - 6 years, 11 months ago

eddie , can you please explain this method ?

Sriram Venkatesan - 6 years, 11 months ago

When they stand in a row no. is 15! Now permuting the 3 in each group is meaningless so we divide by 3! * 3! * 3! * 3! * 3!=6^5. My calculator did the rest.

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