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General rules: find the following values:

C(n,n) = ?

C(n,1) = ?

C(n,0) = ?

Show ALL work, thanks.

2006-12-10 02:16:46 · 4 answers · asked by ÄÐЦÇT¦ÖÑ™ 4 in Science & Mathematics Mathematics

4 answers

C(n,n) = n!/n!(n-n)! = n!/n!0! = n!/n!*1 = 1

C(n,1) = n!/1!(n-1)! = n!/(n-1)! by definition = n

C(n,0) = n!/0!(n-0)! = n!/0!n! = n!/1*n! = n!/n! = 1

2006-12-10 02:30:30 · answer #1 · answered by Modus Operandi 6 · 0 0

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2016-12-11 06:11:59 · answer #2 · answered by ? 4 · 0 0

C(n,n)=n!/[(n-n)!n!]=n!/[0!n!]=n!/n!=1
C(n,1)=n!/[(n-1)!1!]=n(n-1)!/[(n-1)!]=n
C(n,0)=n!/[(n-0)!0!]=n!/n!=1

2006-12-10 05:35:21 · answer #3 · answered by Ranna Renni 2 · 0 0

nCn=n(n-1)(n-2)......(n-r+1)....(5)(4)(3)(2)(1)/
1*2*3*.....(n-4)(n-3)(n-2)(n-1)(n)
=1
another way of looking at it would be nCn=n!/n!(n-r)!
=n!/n!!(n-n)!
=n!/n!0!
=n!/n!
=1
nC1=n things taken 1 at a time so n combinations
or n!/(n-1)!*1!
n(n-1)!/(n-1)!
=n

nC0=1 n things taking0 at a time so only one combination possible
also nC0=nC(n-0)
=nCn
=1
or
n!/(n-0)!*0!
=n!/n!
=1

2006-12-10 03:08:29 · answer #4 · answered by raj 7 · 0 0

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