Topic Included: | Formulas, Definitions & Exmaples. |

Main Topic: | Quantitative Aptitude. |

Quantitative Aptitude Sub-topic: | Permutation and Combination Aptitude Notes & Questions. |

Questions for practice: | 10 Questions & Answers with Solutions. |

**Factorial:** The factorial of an integer is the product of all less than or equal positive integers. It is denoted by \(n!\), where n is any positive integer. The factorial of a negative integer is not possible.$$ n! = n \ (n - 1) \ (n - 2)......3 \times 2 \times 1 $$

**Example:** \(4! = 4 \times 3 \times 2 \times 1 = 24\)

**Example:** \(6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1\) \(= 720\)

**Note:** Remember, \(0! = 1\), \(1! = 1\).

**Permutation:** The arrangement of a number of objects in an ordered manner is called permutation. Here 'order' is important. It is represented by \(nP_r\).$$ nP_r = \frac{n!}{(n - r)!} $$

Where, n = Total number of objects.

P = Permutation.

r = Number of selected objects at a time from the total number of things.

For example let there are 2 Chocolates of different brands for two children P and Q, then Chocolates can be distributed in \(2!\) ways PQ and QP.$$ 2! = 2 \times 1 = 2 \ ways $$

**Example(1):** If there are one Apple, one Banana, and one Orange for three boys P, Q, and R, then three fruits can be distributed in \(3!\) ways.$$ 3! = 3 \times 2 \times 1 = 6 \ ways $$ 6 ways are PQR, PRQ, QPR, QRP, RPQ, RQP.

**Example(2):** If there are 6 Biscuits of different brands for 4 persons, then how many ways Biscuits can be distributed among them?

**Solution:** 6 Biscuits can be distributed among 4 persons by \(6P_4\) ways.$$ nP_r = \frac{n!}{(n - r)!} $$ $$ 6P_4 = \frac{6!}{(6 - 4)!} $$ $$ = \frac{6!}{2!} $$ $$ = \frac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{2 \times 1} $$ $$ = \frac{720}{2} = 360 \ ways $$

**Example(3):** If there are 4 chairs for 3 girls, then how many ways the girls can sit on the chairs?

**Solution:** The girls can sit by \(4P_3\) ways.$$ nP_r = \frac{n!}{(n - r)!} $$ $$ 4P_3 = \frac{4!}{(4 - 3)!} $$ $$ = \frac{4!}{1!} $$ $$ = \frac{4 \times 3 \times 2 \times 1}{1} $$ $$ = \frac{24}{1} = 24 \ ways $$

**Combination:** The selection of a number of objects in which order does not matter is called combination. Here 'order' is not important. It is represented by \(nC_r\).$$ nC_r = \frac{n!}{(n - r)! \times r!} $$

Where, n = Total number of objects.

C = Combination.

r = Number of selected objects at a time from the total number of things.

**Example:** If three students are selected out of 4 students A, B, C, and D, to perform on stage, then how many ways the students can be selected?

**Solution:** The students can be selected by \(4P_3\) ways.$$ nC_r = \frac{n!}{(n - r)! \times r!} $$ $$ 4C_3 = \frac{4!}{(4 - 3)! \times 3!} $$ $$ = \frac{4!}{3!} $$ $$ = \frac{4 \times 3 \times 2 \times 1}{3 \times 2 \times 1} $$ $$ = \frac{24}{6} = 4 \ ways $$ 4 ways are ABC, BCD, CDA, and DAB. Here order does not matter. One thing you have to remember that Permutation is an arrangement whereas Combination is a selection, like in this case, Once three students ABC are selected then they can not be selected again as BCA or ACB, because they all are the same persons. But in the permutation, they can be arranged in three ways ABC, BCA, and ACB.

Lec 1: Permutation and Combination
Lec 2: Binomial Theorem
Exercise-1
Exercise-2
Exercise-3
Exercise-4
Exercise-5