Permutation and Combination Formulas

For those who feel solving Permutation and Combination Problems tough, we have curated simple formulas to make their work easy. You can use them while solving your problems related to the concept and arrive at the solution easily. You will feel the concept of Permutations and Combinations quite easy after referring to the below-outlined Permutation and Combination Formulae List.

1. Factorial Notation

The continuous product of first n natural numbers is called factorial and it can be represented by notation
or n!, n! = 1.2.3 (n- 1).n
n! = n (n – 1)! = n (n- 1) (n – 2)! = n (n- 1) (n – 2) (n – 3)! n(n – 1) ………… (n – r + 1) \(=\frac{n !}{(n-r) !}\)
Note: 0! = 1 and (- n)! = a large number which can not be determined.
Also

2. Fundamental principle of multiplication

Let there are two parts A and B of an operation and if these two parts can be performed in m and n different number of ways respectively, then that operation can be completed in m × n ways.

3. Fundamental principle of addition

If there are two operations such that they can be done independently in m and n ways respectively, then any one of these two operations can be done by (m + n) number of ways.

4. Permutations

• The number of permutations of n different things taken r at a time is ⁿP_r, where ⁿP_r = \(\frac{n !}{(n-r) !}\)
• The number of permutations of n dissimilar things taken all at a time = ⁿP_n = n!
• The number of permutations of n things taken all at a time when p of them are alike and of one kind, q of them are alike and of second kind, r of them are alike and of third kind and all remaining being different is \(\frac{n !}{p ! q ! r !}\)
• The number of ways of n distinct objects taking r of them at a time where any object may be Repeated any number of times is n^r.

5. Restricted Permutations

• The number of permutations of n dissimilar things taken r at a time, when m particular things always occupy definite places = ^n-mP_r-m
• The number of permutations of n different things taken altogether when r particular things are to be placed at some r given places = ^n-rP_n-r = (n – r)!
• The number of permutations of n different things taken r at a time, when m particular things are always to be excluded = ^n-mP_r
• The number of permutations of n different things taken r at a time, when m particular things are always to be included = ^n-mC_r-m × r!

6. Number of Circular Permutations

7. Combinations

The number of combinations of n different things taken r at a time is denoted by ⁿC_r or C (n, r) ⇒ ⁿC_r = \(\frac{n !}{r !(n-r) !}\)

8. Some important results

• ⁿC_r = ⁿC_n-r
• ⁿC_x = ⁿC_y ⇒ x + y = n
• ⁿC_r + ⁿC_r-1 = ⁿ⁺¹C_r
• ⁿC_r = \(\frac{n}{r}\).^n-1C_r-1
• ⁿC_r = \(\frac{1}{r}\) (n – r + 1) ⁿC_r-1
• ⁿC₁ = ⁿC_n-1 = n

9. Restricted combinations

The number of combinations of n different things taking r at a time

• When p particular things are always to be included = ^n-pC_r-p
• When p particular things are always to be excluded = ^n-pC_r
• When p particular things are always included and q particular things are always excluded = ^n-p-qC_r-p

10. Total number of combinations in different cases:

• The number of combinations of n different things taking some or all (or atleast one) at a time = ⁿC₁ + ⁿC₂ + ……… + ⁿC_n = 2ⁿ – 1
• The number of ways to select some or all out of (p + q + r) things where p are alike of first kind, q are alike of second kind and r j are alike of third kind is = (p + 1) (q + 1) (r + 1) – 1
• The number of ways to select some or all out of (p + q + t) things where p are alike of first kind, q are alike of second kind and remaining t are different is = (p + 1) (q + 1)2^t – 1
• The number of combination of r things (r ≤ n) out of n identical things is 1.
• The number of selecting r obejects from n alike objects = n + 1

11. Division into groups

(i) The number of ways in which (p + q) things can be divided into two groups of p and q things is ^p+qC_p = ^p+qC_q = \(\frac{(p+q) !}{p ! q !}\)

(ii) The number of ways in which (p + q + r) things can be divided into three groups containing p, q and r things is
\(\frac{(p+q+r) !}{p ! q ! r !}\)

12. Derangement Theorem

(i) If n items are arranged in a row, then the number of ways in which they can be rearranged so that no one of them occupies the place assigned to it is
\(n !\left[1-\frac{1}{1 !}+\frac{1}{2 !}-\frac{1}{3 !}+\frac{1}{4 !}+\ldots \ldots+(-1)^{n} \frac{1}{n !}\right]\)

(ii) If n things are arranged at n places then the number of ways to rearrange exactly r things at right places is
\(\frac{n !}{r !}\left[1-\frac{1}{1 !}+\frac{1}{2 !}-\frac{1}{3 !}+\frac{1}{4 !}+\ldots . .+(-1)^{n-r} \frac{1}{(n-r) !}\right]\)

13. Some Important results about points

If there are n points in a plane of which m (< n) are collinear, then

• Total number of different straight lines obtained by joining these n points is ⁿC₂ – ^mC₂ + 1
• Total number of different triangles formed by joining these n points is ⁿC₃ – ^mC₃
• Number of diagonals in polygon of n sides is ⁿC₂ – n = \(\frac{n(n-3)}{2}\)
• If m parallel lines in a plane are intersected by a family of other n parallel lines. Then total number of parallelograms so formed is ^mC₂ × ⁿC₂\(\frac{m n(m-1)(n-1)}{4}\)

14. Must learn points

• The coefficitent of x² in the expansion of (1 – x)^-n = ^n-r+1C_r
• If there are l objects of one kind, m objects of another kind, n objects of another kind, then the number of ways of choosing r objects out of these objects i.e. (l + m + n + ………)
• If there are l objects of one kind, m objects of another kind and so on then the no. of way’s of choosing r objects out ot these objects (i.e. l + m + n + …….) is coeff. of x^r in