Table of Contents
Combination in Math:
Each of the various choices or selections which can be formed by taking some or all of a given number of things is called a combination.
Considering the three letters A, B, and C of the English alphabet, the different arrangements or permutations by taking 2 letters at a time are AB, BA, BC, CB, CA, and AC. In permutation, the order must be considered. Thus, AB and BA are different permutations. But in combination, the selections AB and BA are alike (the same objects have been taken into consideration- irrespective of the order in which they have been taken). The order is immaterial in forming combinations. The different types of combinations that can be made by taking two at a time from the letters A, B, and C are AB, BC, and CA.
Number of Combinations of n different things by taking r(n ≥ r) at a time:
Symbolically, it is denoted by nCr or C (n, r). Let x be the required number of combinations in this case. In each combination, there are r things. If the r things in each combination be arranged among themselves in all possible ways, one would get r! permutations in each case.
Thus, from x combinations, we have x . r! permutations. It is also obvious that when the r things in each of the x combinations are arranged among themselves in all possible ways one gets the total number of permutations of n things taken r at a time.
| ∴ x . r! = nPr ⇒ nCr = nPr/r! = n!/[(n – r)! r!] [∵ nPr = n!/[(n – r)!] |
Some Important Results:
| (i) nCr = nCn-r (ii) If nCx = nCy, then x = y or x + y = n (iii) nCr + nCr+1 = n+1Cr+1 (iv) nCr / nCr+1 = (r + 1) / (n – r) (v) nCr = nPr / r! |
| Example 1: Prove that nCr + nCr+1 = n+1Cr+1 Solution- nCr + nCr+1 = n! / [(n – r)! r!] + n! / [n – (r + 1)! (r + 1)!] = n! / [(n – r) (n – r – 1)! r!] + n! / [n – r – 1)! (r + 1) r!] = [n! / {(n – r – 1)! r!}] [(1/(n – r) + 1/(r + 1)] = [n! / {(n – r – 1)! r!}] [(r + 1 + n – r) / {(n – r) (r + 1)}] = [n! (n + 1)] / [(n – r – 1)! r! (n – r) (r + 1)] = (n + 1)! / [{(n + 1) – (r + 1)}! (r + 1)!] = n+1Cr+1 |
| Example 2: If 6P2 = n . 6C2, find n. Solution- 6P2 = n . 6C2 ⇒ 6 x 5 = n [(6 x 5) / (2 x 1)] ⇒ n = 2 Example 3: If nCm = r. nPm, find r. Solution- We know that, nCr = nPr / r! Now, nCm = r. nPm ⇒ nPm / m! = r. nPm ⇒ r = 1/m! |
| Example 4: If nC1, nC2, and nC3 are in A.P., find the value of n. Solution- nC1, nC2, and nC3 are in A.P. ⇒ 2 nC2 = nC1 + nC3 Dividing by nC2 ⇒ 2 = (nC1 / nC2) + (nC3 / nC2) ⇒ 2 = 2/(n – 1) + (n – 2)/3 ⇒ 2 = [6 + (n – 1) (n – 2)] / [3 (n – 1)] ⇒ 2 = (6 + n2 – 2n – n + 2) / (3n – 3) ⇒ 6n – 6 = 6 + n2 – 3n + 2 ⇒ 0 = n2 – 3n + 8 – 6n + 6 ⇒ n2 – 9n + 14 = 0 ⇒ n2 – 7n – 2n + 14 = 0 ⇒ n (n – 7) – 2 (n – 7) = 0 ⇒ (n – 7) (n – 2) = 0 Either n = 7 or n = 2 |
| Example 5: If x = mC2. Prove that xC2 = 3 m+1C4. Solution- x = mC2 ⇒ x = [m (m – 1)]/2 Now, xC2 = [m (m – 1)] / 2C2 ⇒ xC2 = [m (m – 1) / 2] [{m (m – 1) / 2} – 1] / 2 ⇒ xC2 = [m (m – 1) / 2] [{m (m – 1) – 2}/ 2] / 2 ⇒ xC2 = [m (m – 1)] [m (m – 1) – 2] / 8 ⇒ xC2 = [m (m – 1)] [m2 – m – 2] / 8 ⇒ xC2 = [m (m – 1) (m – 2) (m + 1)] / 8 ⇒ xC2 = [(m – 1) (m – 2) m (m + 1)] / 8 ⇒ xC2 = [{(m + 1) (m) (m – 1) (m – 2)} / (4 x 3 x 2 x 1)] x 3 ⇒ xC2 = 3 (m+1C4) |
| Example 6: Evaluate 15C8 + 15C9 – 15C6 – 15C7 Solution- We know that, nCr = nCn-r = 15C8 + 15C9 – 15C6 – 15C7 = 15C7 + 15C6 – 15C6 – 15C7 = 0 Example 7: If x-2P4 = (16/57) (x+2C8), find x. Solution- x-2P4 = (16/57) (x+2C8) ⇒ (x – 2) (x – 3) (x – 4) (x – 5) = (16/57) [(x + 2) (x + 1) (x) (x – 1) (x – 2) (x – 3) (x – 4) (x – 5) / (8 x 7 x 6 x 5 x 4 x 3 x 2 x 1) ⇒ (57 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1) / 16 = (x + 2) (x + 1) (x) (x – 1) ⇒ (21) (20) (19) (18) = (x + 2) (x + 1) (x) (x – 1) ⇒ x – 1 = 18 ⇒ x = 19 |
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