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

by michonamona
Tags: factorial, n or nk, nn1n2nk
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michonamona
#1
Sep20-10, 10:39 PM
P: 122
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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Dick
#2
Sep20-10, 11:05 PM
Sci Advisor
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P: 25,228
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.
Borek
#3
Sep21-10, 02:48 AM
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Adding to what Dick wrote - it may become more obvious when you try to derive the equation.

[tex]\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} [/tex]

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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