This document discusses basics of permutation and combination including: 1) What is factorial and how it is calculated. Factorial of 5 is written as 5! and means multiplying 5 by all lower numbers. 2) What is a permutation. A permutation takes order into account like arrangements of letters A, B, C. 3) What is a combination. A combination does not consider order like the combination of letters A, B, C.

1. BASICS OF PERMUTATION AND COMBINATION by : DR. T.K. JAIN AFTERSCHO ☺ OL centre for social entrepreneurship sivakamu veterinary hospital road bikaner 334001 rajasthan, india www.afterschoool.tk mobile : 91+9414430763

2. What is factorial ? 5! = 5*4*3*2*1 thus factorial 5 means we will multiply 5 to all the numbers lower to it.

3. What is permutation? Permutation means number of arrangements (keeping the order into mind). Thus A,B,C can give you following permutations : ABC,ACB,BAC,BCA,BAC,CAB,CBA

4. What is combination? In combination, the order it not considered, thus A,B,C will have only one combination ABC or BAC or BCA (whatever you call, it is one combination) combination means together, thus when some digits are together, it is a combination, it is not importnat who comes first or second or third.

5. How many numbers can you make out of 4 different digits? Answer : 4! remember, we have to solve it as N!, thus if we have four digits 1,2,3,4, we can make 24 numbers out of it. Verify it : 1234,1243,1324,1342,1423,1432,2134,2143, ...so on

6. How many permutations are possible from 4 digits taking 3 at a time? Total digits = N = 4 digits taken = r = 3 N! / (N-r)! =4! / (4-3)! = 24 answer

7. You have beads of 5 colours, how many necklace can you form? For such questions, use formula of circular permutation = ½ * (n-1)! =1/2 * (5-1)! =12 answer

8. How many combinations can be formed from 5 digits taking 3 at a time? Formula of combination : n! / (r! (n-r)!) n = 5 r=2 5!/(2! *3!) =120/12 =10 answer

9. How many words can you form from C O L L E C T I O N THERE ARE 10 DIGITS. Some are similar (L, C, O, are coming twice). For such the formula is : N! / (X!Y!Z!) N = 10, X = 2, Y = 2, Z = 2 (X, Y and Z are for number of repeatitions) = 10!/ (2! * 2!* 2!)

10. How many combinations are possible out of 6 boys, taking one or all or some of them together? Formula for taking one or all or some of them is : 2^n – 1 here we can take 1, 2, 3, 4,5, or 6 boys at a time. We cannot take 0 boys. So the formula is 2^n – 1 = 2^6 – 1 = 63 answer

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