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Apr 8, 2017

PERMUTATIONS AND COMBINATIONS

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PERMUTATIONS AND COMBINATIONS
Quantitative Aptitude Study Notes for Bank & SSC Exam
PERMUTATIONS AND COMBINATIONS
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 npr denotes the no. of permutations of n different things taken r at a time. The letter P stands for permutation.
Also,
npr = n! / (n-r)!
For example:
No. of permutations made with letters a, b, c by taking two at a time = 3p2 = 3!/1! = 6ways
Six arrangements are given by ab, ba, ac, ca, bc, cb
Notes:
npn = n!,
np0 = 1
total no. of words ( with or without meaning) that can be formed form n distinct letter word is given by npn = n!
Example:
Total no. of words (meaning or without meaning) formed from a word ANIL taking all at a time is given by 4p4 = 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 nCr denotes the no. of combinations of n different things taken r at a time. The letter C stands for combination. Also,
nCr = n! / r!(n – r)!
Example: no. of combinations made with letters a, b, c by taking two at a time =
3C2 = 3! / 2!1! = 3 ways
Three ways are given by ab, bc, ca
Note:
nCn = 1
nC0 = 1
nCr = nCn-r
nPr = r! nCr
no. of ways of arranging n peoples in circular track = (n-1)!
No. of squares in a square of n * n side = 12 +  22 +  32 + ......... n2 = n(n+1)(2n+1) / 6
No. of rectangles in a square of n * n side = 13 +  23 +  33 + ......... n3 = n(n+1)2 / 4
Examples for permutations and Combinations with solution:
Ex. 1: If nP3 = 210, find n.
Solution: n! / (n-3)! = 210
Or, n *  (n-1) * (n-2) = 7 * 6 * 5
n = 7
Ex. 2: the value of 75P2 is:
Solution: 75! / (75 – 2)!
= 75! / 73!
= 75*74*73! / 73!
= 5550
Ex. 3: if nC2 = nC5 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 = 8C4 = 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
= 42C2 = 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
= 16C2 = 16! / 2!(16 – 2)!
= 16 * 15 * 14! / 2 * 1 * 14
= 8 * 15
= 120
Maximum no. of handshakes = 861 + 120 = 981.

To view quantitative study notes on other topics CLICK HERE

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