# Binomial Theorem Aptitude Important Formulas, Definitions, & Examples:

#### Overview:

 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, $$= (5 - 1)!$$ $$= 4! = 4 \times 3 \times 2 \times 1$$ $$= 24 \ ways$$