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Suppose that 10 fish are caught at a lake that contains 5 distinct types of fish. How many different outcomes are possible when at least 2 of the 10 are trout?

My approach:

x1 + x2 + x3 + x4 + x5 = 10

Total number of possible outcomes = (14C4) = 1001

Suppose x5 ≥ 2, then x1 + x2 + x3 + x4 ≤ 8.

If x5 = 0, x1 + x2 + x3 + x4 + x5 = 10, where xi ≥ 0 for i=1, 2, ..., 4. Therefore, there are (13C3) = 286 possible outcomes.

If x5 = 1, x1 + x2 + x3 + x4 + x5 = 9, where xi ≥ 0 for i=1, 2, ..., 4. Therefore, there are (12C3) = 220 possible outcomes.

Thus, number of possible outcomes when at least 2 of the 10 are trout = 1001 - 286 - 220 = 495.

Is my approach correct? Am I double-counting by considering the 2 complement cases?

2 Answers 2

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Actually, you just wish to count the non-negative integer solutions to $x_1+x_2+x_3+x_4+(y_5+2)=10$.

$(y_5+2)$ counts the trout; because we know there are at least two.

So, $x_1+x_2+x_3+x_4+y_5=8$, can be so satisfied in how many ways?

$$\binom{8+4}{4}=495$$

Confirming your answer.

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I make it that this is how many ways can you choose 8 of any 5 different fish, which is stars and bars, $_{12}C_4 = 495$

you have already chosen 2 of the 10 fish which are trout - you can choose more trout.

your approach gets my answer though - do you see my reasoning? After two of the choices are trout, the way it varies is in the 8 remaining choices from 5 (it would be choose from 4 if it was EXACTLY 2 trout) - that is then the classic stars and bars method