Pascal's rule: restrictions on n and k.

  • Thread starter Rasalhague
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According to Wikipedia: Pascal's rule, C(n,k)+C(n,k-1)=C(n+1,k) applies when 0<k<=n+1. But this page says it only applies when 0<k<n. Wikipedia's proof of this version of Pascal's rule involves multiplication by k/k, and by (n+1-k)/(n+1-k). What, if anything, prevents k=n?

Reading on, Corwin seems to only take care to avoid the case where k is strictly greater than n: "In our sum, this means we need to split out the k=0 and k=n+1 terms before applying Pascal's identity." (I've standardised his labelling of variables in this quote.)
 
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According to Wikipedia: Pascal's rule, C(n,k)+C(n,k-1)=C(n+1,k) applies when 0<k<=n+1. But this page says it only applies when 0<k<n. Wikipedia's proof of this version of Pascal's rule involves multiplication by k/k, and by (n+1-k)/(n+1-k). What, if anything, prevents k=n?

Reading on, Corwin seems to only take care to avoid the case where k is strictly greater than n: "In our sum, this means we need to split out the k=0 and k=n+1 terms before applying Pascal's identity." (I've standardised his labelling of variables in this quote.)
Hi Rasalhague! :smile:

The theorem is perfectly valid for k=n. In fact:

C(n,n)=1
C(n,n-1)=n
C(n+1,n)=n+1

Thus C(n,n)+C(n,n-1)=C(n+1,n).
The theorem is even true for k=n+1, provided we define C(n,n+1)=0.
 
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Hi micromass - ever ready to spring to my aid! Then C(n+1,n)=C(n,k)+C(n,k-1) is good as long as k>0, and C(n,k)=C(n-1,k)+C(n-1,k-1) as long as n>0 and k>0.
 
21,992
3,272
Hi micromass - ever ready to spring to my aid! Then C(n+1,n)=C(n,k)+C(n,k-1) is good as long as k>0, and C(n,k)=C(n-1,k)+C(n-1,k-1) as long as n>0 and k>0.
Yep! We can even extend it a bit more if we define C(n,-1)=0 and stuff, but let's not make it even more complicated :smile:
 

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