CBSE Class 10 Mathematics Permutations and Combinations Notes Set 11

Fundamental Principles of Counting

Multiplication principle: if one task can be done in m ways and, for each of these, a second task in n ways, the two together can be done in m × n ways. Addition principle: if two tasks cannot happen together and can be done in m and n ways, then either one can be done in m + n ways.

Factorials and Permutations

n! = n(n − 1)(n − 2)...1, with 0! = 1. A permutation is an ordered arrangement: ⁿPᵣ = n!/(n − r)!. n distinct objects can be arranged in a row in n! ways.
If p objects are alike of one kind, q of another and r of a third, the arrangements of all n objects number n!/(p! q! r!). Arrangements around a circle: (n − 1)!.

Combinations

A combination is a selection where order does not matter: ⁿCᵣ = n!/[r!(n − r)!], with ⁿCᵣ = ⁿCₙ₋ᵣ and ⁿCᵣ + ⁿCᵣ₋₁ = ⁿ⁺¹Cᵣ. The number of selections of at least one object from n distinct objects is 2ⁿ − 1.
Also ⁿPᵣ = ⁿCᵣ × r!: a selection followed by an arrangement.

Common Techniques

Together: glue the objects into one block and multiply by the internal arrangements. Never together: total arrangements minus arrangements with them together, or place the other objects first and put these in the gaps. Digit problems: the first digit cannot be 0, and divisibility tests restrict the last digits. Geometry counts: points on a line give only one line or no triangle.

Solved Examples

Example 1. How many 3-digit numbers can be formed from 1, 2, 3, 4, 5 without repetition?
Answer:
5 × 4 × 3.
60

Example 2. Find the value of ⁸C₃.
Answer:
8 × 7 × 6/(3 × 2 × 1).
56

Example 3. In how many ways can the letters of the word APPLE be arranged?
Answer:
5!/2! because P repeats twice.
60

Example 4. In how many ways can 5 persons be seated around a circular table?
Answer:
(5 − 1)!
24

Example 5. A committee of 3 is to be chosen from 6 people. In how many ways?
Answer:
⁶C₃ = 20.
20

Example 6. In how many ways can 4 boys and 3 girls be arranged in a row so that all girls are together?
Answer:
Treat the girls as one block: 5! ways, times 3! inside the block.
5! × 3! = 720

Example 7. Find n if ⁿP₂ = 56.
Answer:
n(n − 1) = 56, so n = 8.
n = 8

Example 8. How many diagonals does a hexagon have?
Answer:
⁶C₂ − 6 = 15 − 6.
9

Example 9. How many selections of at least one fruit can be made from 5 different fruits?
Answer:
2⁵ − 1.
31

Example 10. How many even 3-digit numbers can be formed from 1, 2, 3, 4, 5 without repetition?
Answer:
The last digit is 2 or 4 (2 ways); the other two places: 4 × 3 = 12.
24