Aptitude Permutations Theory

Formulas

1.Factorial:

Factorial can be defined as

n! = n(n - 1)(n - 2) ...1

2.Permutations:

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

$^nP_{r}$ =$\dfrac{n!}{\left(n-r\right)!}$ = n(n - 1)(n - 2) ... (n - r + 1)

ii.Number of permutations of n distinct things taking them all at a time $^nP_{n}$= n!

iii.$^nP_{0}$= 1

3.Combinations:

i.The number of all combinations of n things, taken r at a time is:

$^nC_{r}$ =$ \dfrac{n!}{\left(r!\right)\left(n-r\right)!}$ = $\dfrac{n\left(n-1\right)\left(n-2\right)...\left(n-r+1\right)}{r!}$

ii. $^nC_{n}$ = 1 and $^nC_{0}$ = 1

iii. $^nC_{r}$ = $^nC_{n-r}$

Permutations and Combinations:

1.Factorial:

Let n be a positive integer. Then n factorial can be defined as

n! = n(n - 1)(n - 2) ...1

Example:

i.0! = 1

ii.1!=1

iii.4! = (4 x 3 x 2 x 1) = 24

iv.5! = (5 x 4 x 3 x 2 x 1) = 120

Exercise:

2.Permutations:

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

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

ii.All permutations made with the letters a, b, c taking all at a time are: (abc, acb, bac, bca, cab, cba)

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

$^nP_{r}$ =$\dfrac{n!}{\left(n-r\right)!}$ = n(n - 1)(n - 2) ... (n - r + 1)

Example:

i.$^6P_{2}$ = (6 x 5) = 30.

ii.$^7P_{3}$ = (7 x 6 x 5) = 210

iii.Number of permutations of n distinct things taking them all at a time $^nP_{n}$= n!

iv.$^nP_{0}$= 1

Exercise:

Example:

How many arrangements can be made out of the letters of the word 'ENGINEERING' ?

The word 'ENGINEERING' has 11 letters.

But in these 11 letters, 'E' occurs 3 times,'N' occurs 3 times, 'G' occurs 2 times, 'I' occurs 2 times and rest of the letters are different.

Hence,number of ways to arrange these letters

=$\dfrac{11!}{\left(3!\right)\left(3!\right)\left(2!\right)\left(2!\right)}$

=$\dfrac{11\times10\times9\times8\times7\times6\times5\times4\times3\times2\times1}{\left(3\times2\times1\right)\left(3\times2\times1\right)\left (2\times1\right)\left(2\times1\right)}$

=277200

Exercise:

3.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

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

Note: AB and BA represent the same selection.

ii.All the combinations formed by a, b, c taking ab, bc, ca.

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

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

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

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

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

$^nC_{r}$ =$\dfrac{n!}{\left(r!\right)\left(n-r\right)!}$ = $\dfrac{n\left(n-1\right)\left(n-2\right)...\left(n-r+1\right)}{r!}$

i. $^nC_{n}$ = 1 and $^nC_{0}$ = 1

ii. $^nC_{r}$ = $^nC_{n-r}$

Example:

i.$^{11}C_{4}$ = $\dfrac{\left(11\times10\times9\times8\right)}{\left(4\times3\times2\times1\right)} = 330$

ii.$^{16}C_{13}$ = $^{16}C_{16-13}$ = $^{16}C_{3}$ = $\dfrac{\left(16\times15\times14\right)}{\left(3\times2\times1\right)} = 560$

Exercise:

select * from question where id =44364;

Example:

How many ways are there in selecting 5 members from 6 males and 5 females, consisting 3 males and 2 females?

This is a case of combination i.e.selecting 3 males from 6 males and 2 females from 5 females.

=Required number of ways = $(6C3 \times 5C2)$

=$\dfrac{6\times5\times4}{3\times2\times1}\times\dfrac{5\times4}{2\times1}$

=200

Exercise:

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