## Permutation and Combination: Definition, Formula, Tricks , Examples

Permutation and Combination General Questions – This is the aptitude questions and answers section on ‘Permutations and Combinations’ with solutions and detailed explanation

## Difference between Permutation and Combination

### Permutation Formula

If the total number of data is “n” and the choice is of “r” things, then permutation will be (without replacement and regarding an order)-

^{n}P_{r} = (n!) / (n-r)!

### Combination Formula

From a group of “n” data, the selection of “r” things without regarding order and replacement-

_{n}C_{r} = (^{n}_{r}) = _{n}P_{r }/ r! = n! / {r! (n-r)!}

These are the key formulas to find out probability permutations and combinations. Moreover, the relation between these two is _{n}C_{r }= ^{n}P_{r} / r!.

Now, let us solve some permutation and combination questions to clear out your doubts.

### Permutation and Combination Formula

*Factorial Notation:* Let *n* be a positive integer. Then, factorial *n*, denoted *n*! is defined as:*n! = n(n – 1)(n – 2) … 3.2.1.Examples:*

We define *0! = 1*.

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

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

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

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

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:

^{n}P_{r} = *n*(*n* – 1)(*n* – 2) … (*n* – *r* + 1) =*n*!/(*n* – *r*)! *Examples:*

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

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

*Cor. number of all permutations of n things, taken all at a time = n!.*

*An Important Result:* If there are *n* subjects of which *p*_{1} are alike of one kind; *p*_{2} are alike of another kind; *p*_{3} are alike of third kind and so on and *p*_{r} are alike of *r*^{th} kind,

such that (*p*_{1} + *p*_{2} + … *p*_{r}) = *n*.Then, number of permutations of these *n* objects is =*n*! / {(*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*. *Examples:*

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.

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

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

Various groups of 2 out of four persons A, B, C, D are:*AB, AC, AD, BC, BD, CD*.

Note that *ab* *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: ^{n}C_{r} =*n*! / { (*r*!)(*n* – *r*)!} = *n*(*n* – 1)(*n* – 2) … to *r* factors. / *r*! *Note:*

^{n}C_{n} = 1 and ^{n}C_{0} = 1.

^{n}C_{r} = ^{n}C_{(n – r)}

### Permutation and Combination Examples

By solving the following permutation and combination problems, you can understand how to derive these formulas for permutation and combination NCERT solutions.

- How to calculate the number of combination and permutation if n = 14 and r = 3

Class 11 permutation and combination solutions:

As per the question, n = 14

r = 3

By deriving the permutation formula-

^{n}P_{r} = (n!) / (n-r)! = 14! / (14 – 3)! = 14! / 11! = (14 X 13 X 12 X 11!) / 11! = 2184

Now, from the combination formula-

_{n}C_{r} = (^{n}_{r}) = _{n}P_{r }/ r! = n! / {r! (n-r)!} = 14! / 3! (14 – 3)! = 14! / 3! (11!) = 14 X 13 X 12 X 11! / 2! X 11!

- 1092

- How many 4- digit numbers can you form from 1, 2, 3, and 4 –

- With repetition?
- Without repetition?

### NCERT Permutation and Combination Solution:

As there will be a 4-digit number, then let the digit be ABCD. Here, D is the unit place, C is the 10^{th} place, B is 100^{th} place, and A is at thousand place.

- Now, with repetition, at the place of D, the possible numbers of the digit are 4. Also, at the

place of A, B, and C, the probable number of digits are 5.

So, the total possible 4-digit numbers are – 4 X 4 X 4 X 4 = 256

- The possible number of digits at the place of D is 4; hence it is the unit place. Now,

without repetition, one digit is occupied at D. So, for place C the possible digit will be 3 and there will be 2 possible digits for B and 1 for A.

Hence, the total possible 4-digit numbers without repetition are – 4 X 3 X 2 X 1 = 24.

From the above permutation and combination questions with solution, you must have understood the pattern of questions which can come in your examinations.