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Let $x,y,z,n$ be positive integers such that $x\leq y\leq z\leq n$. Prove (by counting in two different ways) that:

$\binom {n} {x} \binom {n-x} {y-x} \binom {n-y} {z-y} = \binom {n} {z} \binom {z} {y} \binom {y} {x}$

This is what I have done so far:

Consider an example that there is a box with $n$ beads:

total no. of beads = $n$

No. of square beads = $z$

No. of square beads that are black $=y$

No. of square beads that are black and blue $= x$

Counting method 1: The number of ways to finding an black and blue bead = $\binom {n} {z} \binom {z} {y} \binom {y} {x}$

How do I finish this proof?

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Suppose a coach wants to select a team of z people from a group of n people, and then choose y starters on the team and x captains from among the starters.

If the coach first chooses the team, then the starters from the team, and then the captains from the starters, there are $\displaystyle\binom{n}{z}\binom{z}{y}\binom{y}{x}$ possibilities.

If the coach chooses the captains first, then chooses the remaining starters, and then chooses the rest of the team, there are $\displaystyle\binom{n}{x}\binom{n-x}{y-x}\binom{n-y}{z-y}$ possibilities.

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Ok I have an answer but its not well written:

n-x =>no. of square beads that are not black and blue y-x =>no. of square beads that are black n-y =>no. of beads that are not black z-y =>no.of square beads that are not black

The other counting method should be clear using this.

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