Pascal's rule: restrictions on n and k.

[Rasalhague]

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.)

 

[micromass]

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!

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.

 

[Rasalhague]

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.

 

[micromass]

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

 

[CellCoree]

Based on my understanding of Pascal's rule, it is a mathematical formula used to calculate the number of combinations of k objects from a set of n objects. The restrictions on n and k are necessary in order for the formula to be valid and provide accurate results. If k is equal to n, then the formula would result in a division by zero error, as seen in the Wikipedia proof.

Furthermore, the restriction of 0<k<n is also necessary to ensure that the formula is used for calculating combinations, rather than permutations. If k equals n, then the formula would result in a calculation of n! (n factorial) instead of the desired combination calculation.

In summary, the restrictions on n and k in Pascal's rule are necessary in order for the formula to be valid and provide accurate results for calculating combinations. The proof provided on Wikipedia and the explanation by Corwin both take into account these restrictions to ensure the formula is applied correctly.