Aman Bansal
592 Points
11 years ago

Daer Apurv,

# Combinations - Some important results

1. If out of n different things, any number of items can be chosen, it can be done innC0 +nC1 +nC2 +... +nCn ways. Alternatively, any of n items may or may not be chosen. Hence number of selections = 2×2×...n times = 2n
=> nC0 +nC1 +... +nCn = 2n.
It can be shown that nC0 +nC2 +nC4 +... = nC1 +nC3 +nC4 +... = 2n-1.
2. Out of n different things, at least one (or more) can be chosen in 2n -1 ways.
3. Number of combinations of n different things taken r at a time when p particular things always occur is n -pCr -p. Number of permutations of these is n -pCr -p.r!
4. Number of combinations of n different things taken r at a time when p particular things never occur is n-pCr. Number of permutations of these is n -pCr.r!
5. The number of ways in which m +n different things can be divided into two groups containing m and n things respectively is (m +n)!/[m!n!]. The reason is that whenever you choose a group of m out of m +n, a group of n is automatically left behind. Number of combinations of m +n things taken m at a time is
n +mCm = (m +n)!/[m! (m +n -1)!] = (m + n)!/[m! n!]
6. If subgroups are equal i.e. n = m, then 2m things can be divided into two groups of m each in (2m)!/(m!)² ways. If you distribute things to two persons, then this formula gives number of subdivisions. If it is possible to interchange the two groups then number of divisions is (2m)!/[(m!)².2!]
7. Number of division of m +n +p things into groups of m, n, p things respectively is (m +n +p)!/[m!.n!.p!]
8. If 3m things are divided into 3 equal groups, then number of divisions is (3m)!/(m!)³ and if the groups are interchangeable, the number of divisions is (3m)!/[(m!)³.3!]
9. If there are p +q +r things, where p things are alike, q things are alike and r things are alike, a non-empty selection can be made in (p +1)(q +1)(r +1) -1 ways as 0, 1, 2, ..., p items of p may be chosen; 0, 1, 2, ..., q items of q may be chosen etc.
10. If there are p +q +r things, where p things are alike, q things are alike and remaining r are all different, then a non-empty selection can be made in (p +1)(q +1). 2r -1 ways. (Prove it!)

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