Yahoo Answers: Answers and Comments for Prove nCk = n2Ck + 2(n2Ck1) + n2Ck2
Please explain and show all steps. Thanks!? [Mathematics]
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From Anonymous
enCA
Mon, 28 Mar 2016 14:32:00 +0000
3
Yahoo Answers: Answers and Comments for Prove nCk = n2Ck + 2(n2Ck1) + n2Ck2
Please explain and show all steps. Thanks!? [Mathematics]
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From Nick: This could be considered a double application ...
https://ca.answers.yahoo.com/question/index?qid=20160328143200AAsAfxo
https://ca.answers.yahoo.com/question/index?qid=20160328143200AAsAfxo
Mon, 28 Mar 2016 15:02:22 +0000
This could be considered a double application of Pascal's rule:
C(n,k) = C(n1,k1) + C(n1,k)
= C(n2,k2) + C(n2,k1) + C(n2,k1) + C(n2,k)
= C(n2,k2) + 2C(n2,k1) + C(n2,k)
Also it could be considered a special case of the chuvandermonde identity with i=2:
C(n,k) = C(ni,k)C(i,0) + C(ni,k1)C(i,1) + C(ni,k2)C(i,2) + .. + C(ni,ki+2)C(i,i2) + C(ni,ki+1)C(i,i1) + C(ni,ki)C(i,i)
which counts the ways of choosing k people from ni men and i women. The left hand side counts directly the choices of k people from n and the right counts the choices of k men and 0 women, k1 men and 1 woman, k2 men and 2 women etc.

From Ciel: Yes
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https://ca.answers.yahoo.com/question/index?qid=20160328143200AAsAfxo
Mon, 28 Mar 2016 14:32:11 +0000
Yes