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 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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