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" Theorem "5^(n)P(r)=^(n)C(r)quad r!,0<r...

" Theorem "5^(n)P_(r)=^(n)C_(r)quad r!,0

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If .^(n)P_(r)=x.^(n)C_(r) x=

(n)p_(r)=k^(n)C_(n-r),k=

Assertion (A) : ""_(n)P_(r) gt ""_(n)C_(r) Reason ""_(n)P_(r) = ""_(n)C_(r) xx r!

If ""^(nP_(r)=""^(n)P_(r+1) and ""^(n)C_(r)=""^(n)C_(r-1) find n and r.

If ""^(n)C_(r)=""^(n)C_(r-1) and ""^(n)P_(r+1) = 9 .""^(n)P_(r) then (n,r)=

Prove that ""^(n)P_(r )= ""^(n)C_(r )*^rP_(r ) .

What is the relation between .^(n)P_(r) and .^(n)C_(r) ?

Find n and r if .^(n)P_(r)=336 and^(n)C_(r)=56 .