Suppose that 10 fish are caught at a lake that contains 5 distinct types of fish.

How many different outcomes are possible, where an outcome specifies the numbers of caught fish of each of the 5 types?

How many outcomes are possible when 3 of the 10 fish caught are trout?

How many when at least 2 of the 10 are trout?

Any help is greatly appreciated. I want to find out how to do this without simply using brute force. I came to an answer of 555 for the first question using brute force, but I'm not certain that's correct.

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TopNewestI am unsure whether my answer is correct or not. Someone please comment. But first let me put the full question as it is from its source.

Let me add one more to above (please comment if the solution I wrote for this one is correct or not)

Solution(a)This is star and bar theorem 2 problem. Let us assume there are 10 stars (representing fishes) put in a row. Now we separate them into 5 groups (representing fish types) by putting 4 bars among them. (Notice that if we put bar before first star on extreme left, it will denote that the first group is empty. Same for bar put after last star and two bars put together without any star between them) So we have total 14 positions (10 stars and 4 bars) out of which we have to select 4 for bars which can be done in \(\binom{10+5-1}{5-1}=\binom{14}{4}=4\) ways.(b)When 3 out of 10 are trout, there can be maximum 4 different types of fishes in remaining 7. With same above logic, this can be done in \(\binom{7+4-1}{4-1}=\binom{10}{3}=120\) ways.(c)When at least 2 out of 10 are trout, there can be maximum 5 different types of fishes in remaining 8. This can be done in \(\binom{8+5-1}{5-1}=\binom{12}{4}=495\) ways(d)This becomes star and bar theorem 1 problem. When there is exactly 2 trouts, there should be 4 types of fishes in remaining 8. So not we order 8 stars in a row. They will have 7 gaps separating them. We put 3 bars in those gaps to separate them into 4 groups. This can be done in \(\binom{8-1}{4-1}=\binom{7}{3}=35\) ways. If there are more than 2 trouts, then the remaining 8 fishes must contain 5 types of fishes. This can be done in \(\binom{8-1}{5-1}=\binom{8-1}{5-1}=\binom{7}{4}=35\) ways. By product rule we get \(35\times 35=1225\) – Mahesh Abnave · 4 months, 1 week agoLog in to reply

Are you familiar with stars and bars? – Calvin Lin Staff · 7 months ago

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