Combinations
If n objects are given and we have to choose r (r < n) out of them and the order of choosing these r objects is not important, then such a choice is called a combination of n objects taken r at a time.
^{n}C_{r} = The number of combinations of 'n' things taken 'r' at a time.
Permutations
if n objects are given and we have to arrange r (r < n) out of them and the order of arranging these r objects is important then such a choice is called permutation of n
objects taken r at a time
^{n}P_{r} = = The number of Permutations of 'n' things taken 'r' at a time.
A combination consisting of r objects can give rise to a number of permutation
eg:Q>Find the combination of 4 letters g ,a, t, e, taken 3 at a time
Solution
 g a t
 a t e
 t e g
 e g a
Consider 1.
This combination g a t can give six permutation
 g a t
 g t a
 a g t
 a t g
 t g a
 t a g
So combinations 14 will give rise to 6x4 = 24 permutations
Formulae
 ^{n}C_{r} = =n!/r!(nr)!
 ^{n}P_{r} = =n!/(nr)!
Solved examples
 4 persons enter a railway compartment in which there are 6 seats. In how many ways can they take their places
Solution
4 persons can take 6 seats in ^{6}P_{4} ways
ans = 360 ways
 How many 4 digit numbers can be formed out of the digits 2, 3, 4, 5, 6, 7 if no digit is repeated in any number? How many such numbers will be greater then 4000?
Solution
a) 4 digits can be arranged from 6 digits in ^{6}P_{4} ways
=360 ways
b) Numbers greater than 4000.
Thousands place can be filled in 4 ways (4,5,6,7)
Having filled thousands place now 5 digits remain. So hundreds palce can be filled in ^{5}P_{1} = 5 ways
Having filled hundreds place now 4 digits remain. So tens place can be filled in ^{4}P_{1} = 4ways
Having filled tens place now 3 digits remain. So units place can be filled in ^{3}P_{1} = 3 ways
So the total number of ways in which we can get a number greater than 4000 is 4x5x4x3 = 240 ways
 From 12 books, in how many ways can a selection of 5 be made (a) When one specified book is always included (b)when one specified book is always excluded.
Solution
a) One book is always included. So 4 books can be chosen from the remaining 11 in ^{11}C_{4} ways = 330 ways
b) One book is always excluded. So 5 books can be chosen from the remaining 11 in ^{11}P_{5} ways = 462 ways
 In a party of 15, each person shakes hands with every other person. What is the total number of handshake?
Solution
Consider the first person. He shakes hands with 14 other persons.
1 2,3,4,5,6,7,8,9,10,11,12,13,14,15 = 14 handshakes
2 1,3,4,5,6,7,8,9,10,11,12,13,14,15 . But 2 with 1 is already included above.
So the total number of handshakes = 13
3 1,2,4,5,6,7,8,9,10,11,12,13,14,15. But 3 with 2 and 1 is already included
So the total number of handshakes = 12
so the actual number of handshakes = 14+13+12+11+10+9+8+7+6+5+4+3+2+1 = 105
 There are 8 different locks with exactly one key for each lock. All the keys have been mixed up. What is the maximum no of trials required to find out which key belongs to which lock.
Solution
Consider the first key. At the most all 7 trials will fail .Which implies that the key belongs to the eighth lock
Consider the second key. At the most all 6 trials will fail .Which implies that the key belongs to the seventh lock
So the maximum number of trials will be 7+6+5+4+3+2+1 = 28 trials.
