Factorial : n/(n-k) = n(n-1)(n-2) (n-k+1) - why?

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The equation n!/(n-k)! = n(n-1)(n-2)...(n-k+1) is a mathematical identity that simplifies the calculation of permutations. For instance, with n=4 and k=2, the equation holds true as 4!/2! equals 12, which corresponds to the product of the first two descending integers from n. The right-hand side of the equation stops at (n-k+1), which is crucial for understanding why terms like (n-2) are excluded in specific cases. This shorthand notation effectively captures the essence of permutations without listing all factorial components.

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Why is the equation

(A) n!/(n-k)! = n(n-1)(n-2)...(n-k+1)

true?

For example, let n=4 and k=2, then

4!/2! = 4x3x2x1 / 2x1 = 4x3 = 12.

I understand this example, but I can't make the connection with this and the right-hand-side of equation (A).

For example, why is our example above not

4!/2! = 4(4-1)(4-2)...(4-2+1).

I know this doesn't make any mathematical sense, but I can't understand how the equation on the right-hand-side of (A) is derived.

Thanks for your help.

M
 
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The equation is an informal shorthand. You aren't supposed to include (n-2) as a factor in the case where n=4 and k=2. You are supposed to STOP at (n-k+1)=3.
 
Adding to what Dick wrote - it may become more obvious when you try to derive the equation.

\frac {n!} {(n-k)!} = \frac {n \times (n-1) \times (n-2) \times ... \times (n - k + 1) \times (n - k) \times (n - k -1) \times ... \times 3 \times 2 \times 1} {(n-k) \times (n-k-1) \times (n-k-2) \times ... \times 3 \times 2 \times 1}

Check what cancels out and what is left. And remember that when n and k are too small it is not possible to explicitly list all these terms.
 

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