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. |

**Binomial Theorem:** It is the sum of combinations from zero to \(n\) different objects.$$ nC_0 + nC_1 + nC_2.....nC_n = 2^n $$

**Example:** If there are three balls and a boy is asked to select balls out of three, then find how many ways the boy can select the balls?

**Solution:** If no ball is selected by the boy\(= 3C_0\)

If one ball is selected by the boy\(= 3C_1\)

If two balls are selected by the boy\(= 3C_2\)

If three balls are selected by the boy\(= 3C_3\)

According to the binomial theorem.$$ nC_0 + nC_1 + nC_2.....nC_n = 2^n $$ $$ 3C_0 + 3C_1 + 3C_2 + 3C_3 = 2^3 $$ $$ = 2^3 = 8 \ ways $$

**Important Cases:**

**Case(1):** If there are \(n\) total number of objects and the objects are subdivided into 4 types a, b, c, and d, then the objects can be arranged in \(\frac{n!}{a! \ b! \ c! \ d!}\) ways.

**Example:** How many different words can be formed by using letters of SCHOOL?

**Solution:** In the word SCHOOL, the letter "O" is used 2 times and other letters are used 1 time each, hence $$ = \frac{n!}{a! \ b! \ c! \ d!} $$ $$ = \frac{6!}{2! \ 1! \ 1! \ 1! \ 1!} $$ $$ = \frac{6!}{2!} $$ $$ = \frac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{2 \times 1} $$ $$ = 360 \ words $$

**Case(2):** If there are n total number of objects and have taken all objects at a time, whereas repetition is not allowed then the objects can be arranged in \(n!\) ways.

**Example:** If there are 4 chairs for 4 persons A, B, C, and D, then find how many ways of sitting arrangement are possible?

**Solution:** People can sit in \(4!\) ways.$$ 4! = 4 \times 3 \times 2 \times 1 $$ $$ = 24 \ ways $$

**Case(3):** If there are n total number of objects and have taken \(r\) objects at a time, whereas repetition is allowed then the objects can be arranged in \(n^r\) ways.

**Example:** How many three digit numbers can be formed by 1, 2, 3, and 4 if repetition is allowed?

**Solution:** \(n^r = 4^3 = 64\)

**Case(4): Circular Permutation:** If there are \(n\) number of objects and need to arrange in a circular way, then the objects can be arranged in \((n - 1)!\) ways.

**Note:** In the case of flowers in the garland and beads in a neckless, the flowers or beads can be arranged in \(\frac{(n - 1)!}{2}\) ways.

**Example:** If there are 5 chairs around a dining table in a circular way for five persons, then how many ways of sitting arrangement are possible?

**Solution:** For circular arrangement it can be done in \((n - 1)!\) ways, $$ = (5 - 1)! $$ $$ = 4! = 4 \times 3 \times 2 \times 1 $$ $$ = 24 \ ways $$

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