Question

nCr/nPr=r!

give proof.

Answer 1

Shouldn't it be nPr/nCr = r! ?

Answer 2

yes anonymous is right

Answer 3

agree

Answer 4

nCr = (nPr)/r!

Answer 5

the question is nCr=nPr/r!
give proof.

Answer 6

\[nCr=\frac{n!}{r!(n-r)!}\]
nCr is the number of subsets of r elements taken from a set of n elements.
r! is the number of permutations of a set of r elements. Thus, if we multiply nCr by r!, we get the number of r-permutations of a set of n elements.
In notation:
\[nPr=r!*nCr\]
This probably doesn't count as a proof. This is just an explanation of why this is true.

Answer 7

i never know how much proof is considered proof ...

Answer 8

nPr = n(n-1)(n-2)(n-3)...(n-r+1)

can we use the given defintions or do we have to reinvent the wheel?

Answer 9

\[nPr=\frac{n!}{(n-r)!}\]\[nCr=\frac{n!}{(n-r)!r!}\]
\[nCr = \frac{n!}{(n-r)!r!}*\frac{\frac{1}{(n-r)!}}{\frac{1}{(n-r)!}}=\frac{\frac{n!}{(n-r)!}}{r!}=\frac{nPr}{r!}\]

Answer 10

There is a proof in a book here. It goes like this:
"The r-permutations of a set can be obtained by forming nCr r-combinations of the set and, then, by ordering the elements of each r-combination, which can be done r! ways. Consequently:
nPr = nCr*r!"