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 ⁿC_r= n!/r!(n-r)!

• The relationship between combinations and permutations is ⁿC_r = ⁿP_r/r!

• The number of ways of selecting r objects from n different objects subject to certain condition like:

1. k particular objects are always included =  ^n-kC_r-k

2. k particular objects are never included =  ^n-kC_r

• The number of arrangement of n distinct objects taken r at a time so that k particular objects are

1. Always included = ^n-kC_r-k.r!,

2. Never included = ^n-kC_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-1C_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+1C_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

= ⁿC₁ + ⁿC₂ + ⁿC₃ + ... + ⁿC_n = 2ⁿ - 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 a₁ + a₂ + a₃ + ... a_n + k objects, where a₁ are alike (of one kind), a₂ are alike (of second kind) and so on ... a_n are alike (of nth kind), and k are distinct

= {[(a₁ + 1)(a₂ + 1)(a₃ + 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 2ⁿ – 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 s₁ things from a set of n₁ objects and s₂ things from a set of n₂ objects where combination of s₁ things and s₂ things are independent is given by ^n₁C_s1 x ^n₂C_s2  

• Some results related to ⁿC_r 

1. ⁿC_r = ⁿC_n-r

2. If ⁿC_r = ⁿC_k, then r = k or n-r = k

3. ⁿC_r + ⁿC_r-1 = ⁿ⁺¹C_r

4. ⁿC_r = n/r  ^n-1C_r-1

5. ⁿC_r/ⁿC_r-1 = (n-r+1)/ r

6. If n is even ⁿC_r is greatest for r = n/2

7. If n is odd, is greatest for r = (n-1) /2, (n+1)/2