Permutations and Combinations Formula with Examples for SSC and Bank Exams

5 months ago 976 Views
NEW permutation and combinations formulas

Permutations and combinations formulas are not difficult to use in aptitude questions. You just need to learn the correct way to use formulas and should know how many types of formulas you can use with permutations and combinations equational examples.

With the help of this blog, you can easily learn how to use permutations and combinations formulas with different equational examples. You can increase your score or rank easily in competitive exams with the help of these formulas examples. 

To resolve permutations and combinations problems with solutions click here. As well as, to obtain good marks in this topic you should practice permutation and combination questions.


Formulas of Permutations and Combinations with Examples


Factorial Notation: Let n be positive integer. Then, factorial n,  n! is defined as

n! = n(n – 1)(n – 2)………3.2.1.

Example: (i) 5! = (5×4×3×2×1) = 120; (ii) 4! = (4×3×2×1) = 24 etc.

We define, 0! = 1.

What is permutations?

Permutations: The different arrangement of a given number of things by taking some or all at a time, are called permutations.

Ex.1. All permutation (or arrangements) made with the letters a, b, c, by taking two at a time are (ab, ba, ac, ca, bc, cb).

Ex.2. All permutations made with the letters a, b, c, taking all at a time are : 

(abc, acb, bac, bca, cab, cba).

Number of permutations: Number of all permutations of n things, taken r at a time, is given by:

  $$^nP_r = n(n – 1)(n – 2)………(n-r+1)={n!\over(n-r)!}$$

Example : (i) 6P2 = (6×5) = 30.  (ii) 7P3 = (7×6×5)=210.

Cor. Number of all permutations of n things, taken all at time = n!

An Important Result: If there are n object of which p1 are alike of one kind; p2 are alike of another kind; p3 are alike of third kind and so on and pr are alike of rth kind, such that (p1+p2+pr) = n

Then, number of permutations of these n objects is:

$${n!\over(p_1!).(p_2!)...........(p_r!)} $$

Combinations: Each of the different groups or selections which can be formed by taking some or all of a number of objects, is called a combination.

Ex.1. Suppose we want to select two out of three boys A, B, C. Then, possible selection are AB, BC and CA.

Note that AB and BC and CA.

Note that AB and BA represent the same selection.

Ex.2. All the combination formed by a, b, c taking two at a time are ab, bc, ca.

Ex.3. The only combination that can be formed of three letters a, b, c taken all at a time is abc.

Ex.4. Various groups of 2 out of four persons A, B, C, D are :

AB, AC, AD, BC, BD, CD.

Ex5. Note that ab and ba are two different permutations but they represent the same combination.

Number of Combinations: The number of all combinations of n things, taken r at a time is:

$$^nC_r={n!\over(r!)(n-r)!}={n(n-1)(n-2)....to \ r \ factors.\over r!}$$

  $$ Note \ that\ ^nC_r=1 \ and \ ^nC_0=1$$

An Important Result : nCr = nC(n-r).

$$(i)\ ^{11}C_4 ={(11×10×9×8)\over(4×3×2×1)}=330 $$

  $$(ii)\ ^{16}C_{13}=^{16}C_{(16-13)}=^{16}C_{3}={16×15×14\over3!}={16×15×14\over 3×2×1}=560.$$

Solved

$$Ex.1.\ Evaluate :   {30!\over28!} $$

Explanation

  $$ We\ have, {30!\over28!} ={30×29×(28!)\over28!}=(30×29)=870 $$

Ex.2. Find the value of (i) 60P3 (ii) 4P4

Explanation

  $$(i) \ ^{60}P_{3}={60!\over(60-3)!}={60!\over27!}={60×59×58×(57)!\over57!}=(60×59×58)=205320.$$

(ii) 4P4 = 4! = (4×3×2×1)=24.

Ex.3. Find the value of (i) 10C3 (ii) 50C50

Explanation

$$(i) \ ^{10}C_{3}={10×9×8\over3!}= {10×9×8\over3×2×1}=120$$

$$(iii) ^{50}C_{50}=1.\left( ^{n}C_{n}=1\right)$$

Ex.4. In how many ways can a cricket eleven be chosen out of a batch of 15 players ?

Explanation

(i) Required number of ways =

 $$ \ ^{15}C_{11}=^{15}C_{(15-11)}=^{15}C{4}={15×14×13×12\over 4×3×2×1}= 1365$$

  $$(iii) \  ^{15}C_{11}=\left( ^{n}C_{n}=1\right)$$

Ex.5. In how many ways, a committee of 5 members can be selected from 6 men and 5 ladies, consisting of 3 men and 2 ladies ?

Explanation

(3 men out 6) and (2 ladies out of 5) are to chosen.

∴ Required number of ways = $$\left(^{6}C_{3}×^{5}C_{2}\right)=\left({6×5×4\over3×2×1}×{5×4\over2×1}\right)=200$$

I hope these permutations and combinations formulas are helpful for you. If you have any doubt or want to ask anything related to permutations and combinations formulas with examples, you can ask me in the comment section.

All the best for Exam !

Choose from these tabs.

You may also like

About author

Vikram Singh

Providing knowledgable questions of Reasoning and Aptitude for the competitive exams.

Read more articles

  Report Error : Permutations and Combinations Formula with Examples for SSC and Bank Exams

Please Enter Message
Error Reported Successfully