In today’s blog, we explore the fundamental concepts of Permutation and Combination (P & C) — an essential part of Engineering Mathematics after understanding the basics of counting.

What is Permutation and Combination?

Permutation is the arrangement of items in a particular order.

Combination is the selection of items, where the order does not matter.

Simple Trick to Remember:

• Whenever you only have to select objects, use Combination.
• Whenever you have to select and arrange, use Permutation.

Important: If repetition is allowed in a question, the permutation formula is not applicable; instead, use the Multiplication Principle.

Formulae You Must Know

Permutation (nPr):

$$P_r^n=\frac{n!}{(n–r)!}$$

Combination (nCr):

$$C_r^n=\frac{n!}{r!(n–r)!}$$

Where:

• n = total items
• r = selected items

Where to Use Combination?

Keywords or situations that hint at Combination:

• Formation of a team
• Formation of a group
• Formation of a committee
• Handshakes between people
• Selection of students or employees
• Choosing books or objects

Where to Use Permutation?

Keywords or situations that hint at Permutation:

• Formation of words from letters
• Formation of numbers
• Formation of signals using lights/symbols
• Seating arrangements
• Arranging medals or trophies
• Arranging people in a line

Practice Questions with Answers

Question 1:

How many handshakes are possible between 5 people?

Solution:

Manual Counting:

Handshakes: (P1, P2), (P1, P3), (P1, P4), (P1, P5), (P2, P3), (P2, P4), (P2, P5), (P3, P4), (P3, P5), (P4, P5)

Total = 10 handshakes

Using Formula (Combination):

Select any 2 persons out of 5:

$$C_2^5=\frac{5\times 4}{2}=10$$

Answer: 10 handshakes

Question 2:

If there are 66 handshakes in a party, how many people were there?

Solution:

Let the number of people = n.

Then:

$$C_2^{\mathrm{n}}=\frac{n(n–1)}{2}=66$$

Multiplying both sides by 2:

$$n(n–1)=132$$

Solve by trial or factorization: n = 12 (because 12 × 11 = 132)

Answer: 12 people

Question 3:

How many 4-digit numbers can be formed using digits 1–9?

(a) Without repetition

$$9\times 8\times 7\times 6=3024$$

OR using formula:

$$P_4^9=3024$$

(b) With repetition allowed

Each digit has 10 choices (1 to 9).

$$9\times 9\times 9\times 9=9^4=6561$$

Answer:

• (a) 3024 numbers without repetition
• (b) 6561 numbers with repetition

Question 4:

A committee of 6 is to be selected from 5 boys and 6 girls.

(a) Find the total number of ways to form the committee without any restrictions (i.e., any combination of boys and girls is allowed).

Total people = 11. Select 6 people:

$$C_6^{11}=462$$

(b) Find the number of ways to form the committee such that there are at least 2 girls in the committee

Cases:

• 2 girls and 4 boys: $C_2^6\times C_4^5$ = 15 × 5 = 75
• 3 girls and 3 boys: $C_3^6\times C_3^5$ = 20 × 10 = 200
• 4 girls and 2 boys: $C_4^6\times C_2^5$ = 15 × 10 = 150
• 5 girls and 1 boy: $C_5^6\times C_1^5$ = 6 × 5 = 30
• 6 girls: $C_6^6$ = 1

Adding all: 75 + 200 + 150 + 30 + 1 = 456

Answer: 456 ways

(c) Find the number of ways to form the committee such that there is at least 1 girl in the committee (if committee of 4 people is to be formed)

Total ways without any restriction:

$$C_4^{11}=330$$

Ways with no girl (only boys):

$$C_4^5=5$$

Hence, ways with at least one girl: 330 − 5 = 325

Answer: 325 ways

Question 5:

In how many ways can 5 different books be arranged on 3 shelves?

Solution:

Arrange 3 books from 5:

$$P_3^5=\frac{5!}{(5–3)!}=\frac{5\times 4\times 3}{1}=60$$

Answer: 60 ways

Conclusion

Understanding when to use Combination (selection) and when to use Permutation (selection + arrangement) is key to mastering Engineering Mathematics problems. Remember, if repetition is allowed, you typically apply multiplication instead of standard permutation formulas.