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Theorems


Theorem nPr = nCr r!, 0 r n
The number of combinations of n distinct things taken, 0 r n at a time is
Proof
Let the number of combinations of n things taken r at a time be denoted by x. Take one of these combinations. It contains r things which can be arranged among themselves in r! ways. Therefore, one combination gives rise to r! permutations. Therefore x combinations will give rise to x(r!) permutation. Also the number of permutations of n things taken r at a time is npr.
Thus
                 

Therefore, nPr =nCr r!, 0 r n
Remarks:
  1. The number of ways of selecting no object out of n distinct objects is clearly one, as we do not have to do anything in this case. Therefore nC0 = 1. This could have been derived directly from the formula for nCr. We have
    nC0 =
  2. The number of ways of selecting all the objects from the n distinct objects is 1, since there is only one way to choose all the objects from the n distinct objects. From the formula, we have
    nCn =
  3. The number of ways of selecting r objects out of n distinct objects is equal to the number of ways of selecting (n-r) objects out of n, that is nCr = nCn-r
Proof
Every time we take selection of r objects out of n distinct objects, we are left with a corresponding group of (n-r) objects. Therefore, the number of combinations of n objects taken r at a time is the same as the number of combinations of n objects taken (n-r) at a time.
nCr =,0 r n
nCr === nCr, 0 rn
Hence nCr =nCn-r
  1. If nCa = nCb then a = b or a = n - b
    i.e., n = a + b
Theorem nCr + nCr-1 = n+1Cr
nCr + nCr-1
=+
=+
=+
=+
=
 =n+1Cr




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