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Given 16 Trials equally distributed to 4 individuals ( 4 Trials per person).

How many combinations are there where these conditions are met:
Condition A: 5 Total Successes
Condition B: 2 or Fewer Successes per Individual.

I understand Condition A in exclusivity would be:
$$\binom{16}{5} = 4368$$

I believe Condition B in exclusivity would be the following but please correct me if I am wrong:
$$\left(\binom42 + \binom41 + \binom40\right)^4 = 14641$$

My problem is I do not know how to solve for the number of combinations when both conditions are applied.

Wellwoah
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  • Condition $A$ and $B$ have no restriction on the number of combinations individually. The number of ways to choose 4 individuals on each trial is the same as the number of ways to choose 4 individual on each trial where 5 trials win. It's not until you make it to where an individual can't win three or four times do you actually get a restriction. – ndhanson3 Mar 15 '21 at 05:04
  • Welcome to MSE. Please use MathJax to format your posts. To begin with, surround math expressions (including numbers) with $ signs and use _ for subscripts. $x_1$ comes out as $x_1$. – saulspatz Mar 15 '21 at 05:12

2 Answers2

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There are only two ways to write $5$ as a sum of $4$ nonnegative integers $\leq2$:$$1+1+1+2\\0+1+2+2$$ For the first of these we have $$4\binom41^3\binom42$$possibilities, and for the second we have $$12\binom40\binom41\binom42^2$$ possibilities.

saulspatz
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You have two possibilities for number of trials across $4$ individuals that meet both conditions. Those are $ \{ 2 \ 2 \ 1 \ 0\}$ with permutations $\displaystyle \frac{4!}{2!}$ and $ \{ 2 \ 1 \ 1 \ 1\}$ with permutations $\displaystyle \frac{4!}{3!}$.

So the total number of combinations of trials (considering trials are distinct) meeting both conditions

$ = \displaystyle \frac{4!}{2!} \cdot {4 \choose 2}^2 \cdot {4 \choose 1} + \frac{4!}{3!} \cdot {4 \choose 2} \cdot 4^3 = 3264$

Math Lover
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