There are 10 spaniels, 14 retrievers and 6 poodles at a dog show. 7 dogs are selected to go through to the final.
How many selections of 7 different dogs can be made if there must be at least 1 spaniel, at least 2 retrievers and at least 3 poodles?
Solution
We need to select 7 dogs with at least 1 spaniel, 2 retrievers, and 3 poodles. We consider the following cases:
1 spaniel, 2 retrievers, 4 poodles:
The number of ways to choose 1 spaniel from 10 is \(\binom{10}{1}\).
The number of ways to choose 2 retrievers from 14 is \(\binom{14}{2}\).
The number of ways to choose 4 poodles from 6 is \(\binom{6}{4}\).
Total for this case: \(\binom{10}{1} \times \binom{14}{2} \times \binom{6}{4} = 13650\).
1 spaniel, 3 retrievers, 3 poodles:
The number of ways to choose 1 spaniel from 10 is \(\binom{10}{1}\).
The number of ways to choose 3 retrievers from 14 is \(\binom{14}{3}\).
The number of ways to choose 3 poodles from 6 is \(\binom{6}{3}\).
Total for this case: \(\binom{10}{1} \times \binom{14}{3} \times \binom{6}{3} = 72800\).
2 spaniels, 2 retrievers, 3 poodles:
The number of ways to choose 2 spaniels from 10 is \(\binom{10}{2}\).
The number of ways to choose 2 retrievers from 14 is \(\binom{14}{2}\).
The number of ways to choose 3 poodles from 6 is \(\binom{6}{3}\).
Total for this case: \(\binom{10}{2} \times \binom{14}{2} \times \binom{6}{3} = 81900\).
Adding all cases together gives the total number of selections:
\(13650 + 72800 + 81900 = 168350\).
Alternatively, rounding or different interpretations may lead to 168000.
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