# Permutation and Combination

29 April 2020

In this article, we will discuss the basic concepts and formulas of Permutation & Combination required for solving problems in various placement entrance tests and competitive exams.The questions from this topic are mainly focused on checking the skill of an aspirant in logical counting.

Learn the fundamental principles of Permutation and Combination along with solved examples.

##### Formula's for Permutation and Combination

Factorial Notation:

Let n be a positive integer. Then, factorial n, denoted n! is defined as:

n! = n(n - 1)(n - 2) ... 3.2.1.

Examples:

- We define 0! = 1.

- 4! = (4 x 3 x 2 x 1) = 24.

- 5! = (5 x 4 x 3 x 2 x 1) = 120.

Permutations:

The different arrangements of a given number of things by taking some or all at a time, are called permutations.

Examples:

- All permutations (or arrangements) made with the letters a, b, c by taking two at a time are

(ab, ba, ac, ca, bc, cb).

- All permutations made with the letters a, b, c taking all at a time are:

( abc, acb, bac, bca, cab, cba)

Number of Permutations:

Number of all permutations of n things, taken r at a time, is given by:

nPr    = n! / (n - r)!

Examples:

- 6P2 = (6 x 5) = 30.

- 7P3 = (7 x 6 x 5) = 210.

Combinations:

Each of the different groups or selections which can be formed by taking some or all of a number of objects is called a combination

Examples:

- Suppose we want to select two out of three boys A, B, C. Then, possible selections are AB, BC and CA.

- Note: AB and BA represent the same selection.

- All the combinations formed by a, b, c taking ab, bc, ca.

- The only combination that can be formed of three letters a, b, c taken all at a time is abc.

- Various groups of 2 out of four persons A, B, C, D are:

AB, AC, AD, BC, BD, CD.

- Note that ab ba is two different permutations but they represent the same combination.

Number of Combinations:

The number of all combinations of n things, taken r at a time is:

nCr = n! / (r !) (n - r) !

Note:

nCn = 1 and nC0 = 1.

nCr  = nC(n-r)

Examples:

11C4 = (11 x 10 x 9 x 8) / (4 x 3 x 2 x 1) = 330.

16C13 = 16C(16-13) 16C13  = (16 x 15 x 14) /  (3 x 2 x 1) = 560.

##### Practice Questions For Permutation and Combination

Q1)An intelligence agency forms a code of two distinct digits selected from 0, 1, 2, …., 9 such that the first digit of the code is nonzero. The code, handwritten on a slip, can however potentially create confusion, when read upside down-for example, the code 91 may appear as 16. How many codes are there for which no such confusion can arise?

A) 80

B) 78

C) 71

D) 69

Solution :

Q2) If a refrigerator contains 12 cans such that 7 blue cans and 5 red cans. In how many ways can we remove 8 cans so that at least 1 blue can and 1 red can remains in the refrigerator.

A) 513

B) 455

C) 627

D) 369

Solution :

Q3) There is meeting of 20 delegates that is to be held in a hotel. In how many ways these delegates can be seated along a round table, if three particular delegates always seat together?

A) 17! 3!

B) 18! 3!

C) 17! 4!

D) 18! 4!

Solution :

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Q4) In how many ways a four digit even number can be formed by using the digits 2,3,5,8 exactly once?

A) 24

B) 12

C) 6

D) 18

Solution :

Q5) A college has 10 basketball players. A 5-member team and a captain will be selected out  of these 10 players. How many different selections can be made?

A) 1260

B) 210

C) 10C6 * 6!

D) 10C5 * 6

Solution :

A team of 6 members has to be selected from the 10 players. This can be done in 10C6 or 210 ways.
Now, the captain can be selected from these 6 players in 6 ways.
Therefore, total ways the selection can be made is 210*6 = 1260.

Q6) Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?

A) 24400

B) 21300

C) 210

D) 25200

Solution :

Number of ways of selecting 3 consonants from 7= 7C3
Number of ways of selecting 2 vowels from 4= 4C2
Number of ways of selecting 3 consonants from 7 and 2 vowels from 4
= 7C3 × 4C2
=(7×6×5/3×2×1)×(4×3/2×1)=210
It means we can have 210 groups where each group contains total 5 letters (3 consonants and 2
vowels).
Number of ways of arranging 5 letters among themselves=5!=5×4×3×2×1=120
Hence, required number of ways=210×120=25200

Q7) A boy has nine trousers and 12 shirts. In how many different ways can he select a trouser and a shirt?

A) 21

B) 12

C) 9

D) 108

Solution :

The boy can select one trouser in nine ways.
The boy can select one shirt in 12 ways.
The number of ways in which he can select one trouser and one shirt is 9 * 12 = 108 ways.

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Q8) Find the sum of all the 4 digit numbers that can be formed with the digits 3, 4, 5 and 6

A) 119988

B)  11988

C) 191988

D) None of these

Solution :

No. of Digits = 4 All are distinct; They can be arranged in 4! = 24 ways
Each of the digits 3, 4, 5 and 6 occur at unit place = 3! Ways = 6 ways.
Thus there will be 6 numbers ending with 3, 4, 5 and 6 each. So the sum of the digits at unit's
place = 6(3 + 4 + 5 + 6) =108
The sum of numbers = 108 × 103 + 108 × 102 + 108 × 101 + 108 × 100 = 119988

Q9)In how many different ways can the letters of the word 'LEADING' be arranged in such a way that the vowels always come together?

A) 720

B) 520

C) 700

D) 750

Solution :

The word 'LEADING' has 7 different letters.
When the vowels EAI are always together, they can be supposed to form one letter. Then, we have to arrange the letters LNDG (EAI).
Now, 5 (4 + 1) letters can be arranged in 5! = 120 ways.
The vowels (EAI) can be arranged among themselves in 3! = 6 ways.
Required number of ways = (120 x 6) = 720.

Q10) How many combinations of students are possible if the group is to consist of exactly 3 freshmen?

A) 5000

B) 4550

C) 4000

D) 3550

Solution :

Here we need the number of possible combinations of 3 out of 5 freshmen,5C3, and the number
of possible combinations of 3 out of the 15 sophomores and juniors,
15C3. Note that we want 3 freshmen and 3 students from the other classes. Therefore, we
multiply the number of possible groups of 3 of the 5 freshmen times the number of possible
groups of 3 of the 15 students from the other classes. 5C3×15C3=4,550.

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