Click to Chat
1800-2000-838
+91-120-4616500
CART 0
Use Coupon: CART20 and get 20% off on all online Study Material
Welcome User
OR
LOGIN
Complete Your Registration (Step 2 of 2 )
Revision Notes on Combinations
If certain objects are to be arranged in such a way that the order of objects is not important, then the concept of combinations is used.
The number of combinations of n things taken r (0 < r < n) at a time is given by ^{n}C_{r}= n!/r!(n-r)!
The relationship between combinations and permutations is ^{n}C_{r} = ^{n}P_{r}/r!
The number of ways of selecting r objects from n different objects subject to certain condition like:
k particular objects are always included = ^{n-k}C_{r-k}
k particular objects are never included = ^{n-k}C_{r}
The number of arrangement of n distinct objects taken r at a time so that k particular objects are
Always included = ^{n-k}C_{r-k}.r!,
Never included = ^{n-k}C_{r}.r!.
In order to compute the combination of n distinct items taken r at a time wherein, the chances of occurrence of any item are not fixed and may be one, twice, thrice, …. up to r times is given by ^{n+r-1}C_{r}
If there are m men and n women (m > n) and they have to be seated or accommodated in a row in such a way that no two women sit together then total no. of such arrangements
= ^{m+1}C_{n}. m! This is also termed as the Gap Method.
If there is a problem that requires n number of persons to be accommodated in such a way that a fixed number say ‘p’ are always together, then that particular set of p persons should be treated as one person. Hence, the total number of people in such a case becomes (n-m+1). Therefore, the total number of possible arrangements is (n-m+1)! m! This is also termed as the String Method.
Let there be n types of objects with each type containing at least r objects. Then the number of ways of arranging r objects in a row is n^{r}.
The number of selections from n different objects, taking at least one
= ^{n}C_{1} + ^{n}C_{2} + ^{n}C_{3} + ... + ^{n}C_{n} = 2^{n} - 1.
Total number of selections of zero or more objects from n identical objects is n+1.
Selection when both identical and distinct objects are present:
The number of selections, taking at least one out of a_{1} + a_{2} + a_{3} + ... a_{n} + k objects, where a_{1} are alike (of one kind), a_{2} are alike (of second kind) and so on ... a_{n} are alike (of nth kind), and k are distinct
= {[(a_{1} + 1)(a_{2} + 1)(a_{3} + 1) ... (a_{n} + 1)]2^{k}} - 1.
Combination of n different things taken some or all of n things at a time is given by 2^{n} – 1.
Combination of n things taken some or all at a time when p of the things are alike of one kind, q of the things are alike and of another kind and r of the things are alike of a third kind
= [(p + 1) (q + 1)(r + 1)….] – 1
Combination of selecting s_{1} things from a set of n_{1} objects and s_{2} things from a set of n_{2} objects where combination of s_{1} things and s_{2} things are independent is given by...
Post Question
Dear , Preparing for entrance exams? Register yourself for the free demo class from askiitians.
Solved Examples on Combinations Illustration 1:...