**PERMUTATIONS AND COMBINATIONS**

**Quantitative Aptitude Study Notes for Bank & SSC Exam**

You know that quantitative
aptitude section is most important in

**bank exams**in PO and Clerk and for other competitive exams because if you want good score in bank exam then you have to score good in maths. In competitive exams the most important thing is time management, if you know how to manage your time then you can do well in**Bank Exams**as well as in other competitive exams. That’s where**maths shortcut tricks and formula are**comes into action. So continuously we are providing shortcut tricks on different maths topics. Today’s topic is Permutations and combinations. This is the one of the most important topic in quantitative aptitude section in bank and SSC exam. You should know how to**Permutations and combinations**questions and answers in very short time for bank exam. From this chapter around 1-2 questions are given in the SBI and IBPS exams. For this here we are providing shortcut tricks and quicker method to solve permutations and combinations problems in very short time.**In first we will look at some important points, which are most important for permutations and combinations and also for probability.**

**FACTORIAL NOTATION:**

The product of n consecutive
positive integers beginning with 1 is denoted by n! Is read as factorial n or 5!
is read as factorial five, etc.

So, n! = 1*2*....... * (n-2) *
(n-1) * n

= n * (n – 1) * (n – 2) * ..... *
2 * 1

For example,

5! = 5 * 4 * 3 * 2 * 1 = 120

Or

7! = 7 * 6* 5 * 4 * 3 * 2 * 1 = 5040

So , by this method we can find
out factorial of any number.

Factorial of some no.s are given
below:

**0! = 1**

**1! = 1**

**2! = 2**

**3! = 6**

**4! = 24**

**5! = 120**

**6! = 720**

**7! = 5040**

**8! = 40320**

**9! = 362880**

**10! = 3628800**

**Permutation (Arrangements):**

Each of the different arrangements
which can be made by taking some or all of
the given things or objects at time is called a

**permutation.**The symbol^{n}p_{r}denotes the no. of**permutations**of n different things taken r at a time. The letter P stands for permutation.
Also,

^{n}p

_{r }= n! / (n-r)!

**For example:**

No. of permutations made with
letters a, b, c by taking two at a time =

^{3}p_{2 }= 3!/1! = 6ways
Six arrangements are given by ab,
ba, ac, ca, bc, cb

Notes:

^{n}p

_{n }= n!,

^{n}p

_{0 }= 1

total no. of words ( with or without
meaning) that can be formed form n distinct letter word is given by

^{n}p_{n }= n!**Example:**

Total
no. of words (meaning or without meaning) formed from a word ANIL taking all at
a time is given by

^{4}p_{4 }= 4!
Total no. of words (meaning or
without meaning) formed a given word of n letters in which particular letters
are repeated in m, k times is given by = n! / m!k!

**Example:**total no. of words (meaning or without meaning) formed form a word KANCHAN taking all at a time is given by

7!/2!2! as in KANCHAN total no.
of letter are 7 in which A & N is repeated two times each.

**Combinations:**

Each of the different selections
or groups which can be made by taking some or all of a no. of given things or
objects at a time is called a combination. The symbol

^{n}C_{r }denotes the no. of**combinations**of n different things taken r at a time. The letter C stands for combination. Also,^{n}C

_{r }= n! / r!(n – r)!

**Example:**no. of combinations made with letters a, b, c by taking two at a time =

^{3}C

_{2 }= 3! / 2!1! = 3 ways

Three ways are given by ab, bc,
ca

Note:

^{n}C

_{n }= 1

^{n}C

_{0 }= 1

^{n}C

_{r }=

^{n}C

_{n-r }

^{n}P

_{r }= r!

^{n}C

_{r}

no. of ways of arranging n
peoples in circular track = (n-1)!

No. of squares in a square of n *
n side = 1

^{2}+ 2^{2}+ 3^{2}+ ......... n^{2}= n(n+1)(2n+1) / 6
No. of rectangles in a square of
n * n side = 1

^{3}+ 2^{3}+ 3^{3}+ ......... n^{3}= n(n+1)^{2}/ 4**Examples for permutations and Combinations with solution:**

**Ex. 1:**If

^{n}P

_{3 }= 210, find n.

**Solution**: n! / (n-3)! = 210

Or, n * (n-1) * (n-2) = 7 * 6 * 5

n = 7

**Ex. 2:**the value of

^{75}P

_{2 }is:

**Solution:**75! / (75 – 2)!

= 75! / 73!

= 75*74*73! / 73!

= 5550

**Ex. 3**: if

^{n}C

_{2}=

^{n}C

_{5 }find n.

**Solution:**n! / 2!(n – 2)! = n! / 5!(n – 5)!

Or, 5!(n – 5)! = 2!(n – 2)!

Or, 5 * 4 * 3 * 2 * (n – 5)! = 2
* (n – 2) * (n – 3) * (n – 4) * (n – 5)!

Or, 5 * 4 * 3 = (n – 2) * (n – 3)
* (n – 4)

N – 2 = 5

0r

N = 7

**Ex. 4:**how many quadrilaterals can be formed by joining the vertices of an octagon?

**Solution:**A quadrilateral can be formed by joining the vertices of an octagon?

**Solution:**A quadrilateral has 4 sides or 4 vertices whereas an octagon has 8 sides or vertices.

Reqd no. of quadrilateral =

^{8}C_{4 }= 8! / 4!(8 – 4)!
= 8 * 7 * 6 * 5 / 24

= 70

**Ex. 5:**with specific rule: Suppose there are 42 men and 16 women in a party. Each man shakes his hand only with all the men and each woman shakes her hand only with all the women. We have to find the maximum no. of handshakes that have taken place at the party.

Solution: from each group of two
persons we have one handshake.

**Case 1: total no. of handshakes among the group of 42 men**

=

^{42}C_{2 }= 42! / 2!(42! – 2!)
= 42! / 2! 40!

= 42 * 41 * 40! / 2 * 1 * 40!

= 21 * 41

= 861

**Case 2: total no. of handshakes among the group of 16 women**

=

^{16}C_{2 }= 16! / 2!(16 – 2)!
= 16 * 15 * 14! / 2 * 1 * 14

= 8 * 15

= 120

Maximum no. of handshakes = 861 +
120 = 981.

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