Permutation
Permutation means arrangement of things.In permutations sequence is important.Formulae for permutations npr = n! / (n-r)!
For example:
In how many different ways 3 letters a, b ,c arranged 2 at a time
6 ways (ab, bc, ca, ba, cb, ac)
In a simple way, 3p2= 3! / (3 − 2)!
= 3 × 2 × 1/1 = 6
We don’t use the formulae all the time because its consumes a lot of time so we can use a small technique
nPr = the product from ‘n’ to ‘r’ numbers
3p2 = 3 × 2 = 6 (product from 3 to 2 numbers)
4p3 =4 × 3 × 2 = 24 (product from 4 to 3 numbers)
Combination
Combination means selection of things. In combinations sequence is not important. n!
Formulae for combinations ncr = n! / (n-r)! r!
For example:
In how many different ways 3 letters a, b, c selected 2 at a time
3 ways (ab, bc, ca)
In a simple way, 3C2= 3!/(3 − 2)!2!
=3 × 2 × 1/2 × 1 = 3
We don’t use the formulae all the time because its consumes a lot of time so we can use a small technique
nCr = the product from n to r numbers / product from r to 1
3C2 = 3 × 2 (product from 3 to 2 numbers) / 2 × 1 (product from 2 to 1) = 3
4C3 = 4 × 3 × 2(product from 4 to 3 numbers) / 3 × 2 × 1 (product from 3 to 1) = 4
Note: If nCx = nCy, then x = y or x + y = n
If we want to calculate value of 5C4 just we can calculate the value of 5C1,
The reason is (x + y = n) means 4 + 1 = 5
As the same way (15C4 = 15C11,11C7 = 11C4 .........)
E.g.: 1. nC2 = nC3 then find n = ?
n = 2 + 3 = 5(x + y = n)
2. 19C3r = 19Cr + 3
Either 3r = r + 3 or 3r + r + 3 = 19
2r = 3 or 3r + r + 3 = 19
r = 3/2 or 4r + 3 = 19
r = 3/2 or 4r = 16 then r = 4
r can’t be fraction so r = 4
Principle of addition: Whenever we find “or” we can do addition
Principle of multiplication: Whenever we find “and” we can do multiplication
Model Questions
1. If 10Pr = 720, find r = ?
A) 3 B) 2 C) 5 D) 6 E) None of these
Explanation: 10Pr = 720
10! /(10 − r)! = 10 × 9 × 8 = 720
Ans: A
2. If nP3 = 60, find n = ?
A) 3 B) 2 C) 5 D) 6 E) None of these
Explanation:nP3 = 60
n! /(n − 3)!= 60
n(n − 1)(n − 2) = 60
5 × 4 × 3 = 60
Ans: C
3. In how many different ways can 5 persons stand in a row for a photograph?
A) 100 B) 120 C) 50 D) 5 E) None of these
Explanation: 5 places we can arrange the 5 people so 5P5 = 5! = 120
Ans: B
4. How many 4 digit numbers can be formed with the digits 1, 2, 3, 4, 5?
A) 100 B) 120 C) 50 D) 5 E) None of these
Explanation: 4 digits we can select from 5 digits so 5P4 = 5 × 4 × 3 × 2 = 120
Ans: B
5. How many different words can be formed using the letters of the word ‘BANKER’?
A) 120 B) 6 C) 720 D) 12 E) None of these
Explanation: 6 letters are arranged in 6! ways 6P6 = 6! = 6 × 5 × 4 × 3 × 2 × 1 = 720
Ans: C
6. How many different ways, can the letter of the word ‘STRESS’ be arranged?(IBPS Clerk 2010)
A) 120 B) 6 C) 720 D) 12 E) None of these
Explanation: 6 letters are arranged in 6! Ways
“S” is repeated 3 times so 6! / 3!
= 6 × 5 × 4 × 3 × 2 ×1 / 3 × 2 × 1= 120
Ans: A
7. How many different ways, can the letter of the word ‘VENTURE’ be arranged? (IBPS Clerk 2011)
A) 1250 B) 600 C) 2520 D) 1200 E) None of these
Explanation: 7 letters are arranged in 7! Ways
“E” is repeated 2 times so 7! / 2!
= 7 × 6 × 5 × 4 × 3 × 2 × 1 / 2 × 1 = 2520
Ans: C
8. how many 4 digit numbers can be formed with digits 0, 1, 3, 6
A) 18 B) 63 C) 24 D) 12 E) None of thes
Explanation: First place we can arrange with 1, 3, 6
If we can arrange with 0 it is not at all a four digit number
Remaining 3 places we can arrange with 3! ways 3 × 3! = 18
Ans: A
9. In how many ways committee of 3 men and 2 women can be formed out of total of 4 men and 4 women?
A) 15 B) 16 C) 20 D) 24 E) None of these
Explanation: A committee of 3 men and 2 women can be formed out of total of 4 men and 4 women
4C3 × 4C2 = 4 × 6 = 24
Ans:D
10. At a meeting, each person gave a handshake to the rest of the people, if the total number of handshakes was 28,find number of people at the meet?
A) 14 B) 9 C) 7 D) 8 E) None of these
Explanation: With the help of 2 persons one shake hand is possible so from ‘n’ persons we can select 2 persons then will know no of shake hands
nC2 = 28
n(n − 1) / 2 = 28
n = 8
Ans:D
11. There are two Urn’s. Urn A has 3 distinct red balls and urn B has 9 distinct blue balls. From each urn two balls are taken out at random and then transferred to the other. The number of ways in which this can be done is?
A) 3 B) 36 C) 66 D) 108 E) None of these
Explanation: In Urn ‘A’ we can choose 2 balls from 3 distinct red balls and urn ‘B’ we can choose 2 balls from 9 distinct blue balls.
3C2 × 9C2 = 3 × 36 = 108
Ans: D
Directions (Qs. 12 - 17): Different committees are to be made as per requirement given in each question, in how many different ways can it be done?
(IBPS PO 2012)
8 students out of which 5 are doctors and 3 are scientists.
12. A committee of 4 in which 3 are doctors and 1 is scientist
A) 15 B) 16 C) 20 D) 30 E) None of these
Explanation: A committee of 4 in which we have to select 3 doctors from 5 doctors and 1 scientist selected from 3 scientists
5C3 × 3C1 = 10 × 3 = 30
Ans: D
13. A committee of 5 in which 3 are doctors.
A) 15 B) 16 C) 30 D) 24 E) None of these
Explanation: A committee of 5 members in which we have to select 3 are doctors, so the remaining 2 we can select fromscientists of 3.
5C3 × 3C2 = 10 × 3 = 30
Ans: C
14. A committee of 2 in which there is no doctors.
A) 5 B) 6 C) 2 D) 3 E) None of these
Explanation: A committee of 2 in which there is no doctors so we can select the 2 members from 2 scientists 3C2 = 3
Ans: D
15. A committee of 3 in which there is no scientist.
A) 10 B) 16 C) 20 D) 24 E) None of these
Explanation: A committee of 3 in which there is no scientist, so we can select the 3 members from 5 doctors.
5C3 = 5C2 = 10
Ans:A
16. A committee of 2 in which either both are doctors (or) both are scientists.
A) 15 B) 13 C) 20 D) 24 E) None of these
Explanation: A committee of 2 either both are doctors or both are scientists. So we can select 2 doctors from 5 doctors or 2 scientists from 3 scientists
5C2 + 3C2 = 10 + 3 = 13
Ans: B
17. A committee of 3 in which at least one doctor is there.
A) 15 B) 55 C) 20 D) 24 E) None of these
Explanation: A committee of 3 at least 1 doctor so we can select 1 doctor and 2 scientists or 2 doctors an 1 scientist or doctors 3 from 5 doctors and 3 scientists.
5C1 × 3C2 + 5C2 × 3C1 + 5C3 = 55
Ans: B
