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Applications Of Permutation - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

Quick Facts

  • APPLICATION OF PERMUTATION-I is considered one of the most asked concept.

  • 173 Questions around this concept.

Solve by difficulty

At an election, a voter may vote for any number of candidates, not greater than the number to be elected. There are 10 candidates and 4 are to be elected. If a voter votes for at least one candidate, then the number of ways in which he can vote is:

How many different four-letter words can be formed using the 10 alphabetic letters without repetition, if the word must start with the letter "C" and end with the letter "E", while the second and third letters can be any of the remaining three letters?

 

How many different six-letter words can be formed using 10 reverse alphabetic letters as first letter from Z without repetition?

 

In how many ways can the letters of the word "ENGINEER" be arranged such that the N and E should not be together?

If n is the number of ways five different employees can sit into four indistinguishable offices where any office may have any number of persons including zero, then n is equal to :

There are three men have 3 coats,4 waist coats and 6 caps. In how many ways can they wear them? Also ensure that no 3 men should wear both caps and coats?

 

Concepts Covered - 6

APPLICATION OF PERMUTATION-I

Arrangement of Digits (Part 1)

To make 5 digit number from 1,2,3,4,5,6,7

    1. If the repetition of digits is not allowed

    In this case,

    First place of digit (from left) can be filled in 7 ways

    Second place can be filled in 6 ways (as one number is already used at first place)

    Third place can be filled in 5 ways

    Fourth in 4 ways

    Fifth in 3 ways

    Hence the total number of ways is 7 x 6 x 5 x 4 x 3 = 2520.

    2. If the repetition of digits is allowed

    In this case,

    First place can be filled in 7 ways

    Second place can be filled in 7 ways (as repetition is allowed)

    Similarly all other places can be filled in 7 ways

    Hence the total number of ways is 7 x 7 x 7  x 7 x 7 = 75.

    3. If the number is EVEN

    In this case,

    One's place can be filled by three numbers (2,4,6) as even number can only end with (0,2,4,6,8). Hence we have three different possibilities.

    If one's place is occupied by 2 then remaining 6 digits can be arranged in remaining 4 places in ^6P_4 ways

    OR,

    If one's place is occupied by 4 hence remaining can be arranged in ^6P_4 ways

    OR,

    If one's place is occupied by 6 hence remaining can be arranged in ^6P_4 ways

    Hence the total possibilities are ^6P_4+^6P_4+^6P_4=1080

 

Let us take another example

Using 0,1,2,3,4,5 how many 4-digit numbers can be formed if the repetition of digits is not allowed.

Here 0 cannot be at a leading place as in such a case the number will become a 3-digit number.
Hence the first place can be filled in 5 ways and the second place left with 5 possibilities (now we have one lesser digit available as it is used up at leading place, but we have one more digit '0' that can be filled at second place) that's why it can also be filled with 5 ways and the third-place left with 4 possibilities and the fourth place left with 3 possibilities.

So the total number of 4 digit numbers = 5×5×4×3=300

Solving the same question if repetition of digits was allowed

In that case, the first place (from left) has 5 choices (excluding 0 otherwise it will not be 4 digit number). 2nd, 3rd, and 4th place will have 6 choices (all choices available, repetition allowed), so the total number, in this case, = 5×6×6×6

APPLICATION OF PERMUTATION-II

Arrangements of Digits (Part 2)

Find the number of Distinct 4-digits numbers that can be formed using 1,2,3,4,5,6,7 for each of the following condition:

1. If the number is greater than 3000.

The thousand’s digit can be filled in 5 ways (any one of 3, 4, 5, 6 or 7). Now one is left with 6 digits and has to arrange 3 of them. This can be done in 6P3 = 6 × 5 × 4 ways. Thus a total of 5 × (6 × 5 × 4) = 600 such numbers can be formed.

2. If the number is EVEN.

The unit’s digit can be filled in three ways (2, 4 or 6). Now we are left with 6 digits and have to arrange 3 of them. This can be done in 6P3 = 6 × 5 × 4 ways. Thus a total of (6 × 5 × 4) × 3 = 360 such numbers can be formed.

3. If the number is EVEN and greater than 3000.

The unit digit could be 2 OR 4 OR 6

Considering the case that the units digit is 2:

The units digit can be filled in only 1 way, with a 2. Having filled this, all of 3, 4, 5, 6 and 7 are available (so that the number is greater than 3000) for the thousands place and thus it can be filled in 5 ways (3,4,5,6,7). Next, we are left with 5 digits and two of which have to be arranged i.e. can be done in 5 × 4 ways. Thus total possible numbers with units digit being 2 will be 5 × (5 × 4) × 1 = 100

Considering the case that the units digit is 4: The units digit can be filled in 1 way. Having filled this, the digits available for the thousands place are 3, 5, 6, 7 i.e. there are further 4 possibilities. Next, we are left with 5 digits and two of which have to be arranged i.e. can be done in 5 × 4 ways. Thus total possible numbers with unit’s digit being 4 or 6 will be 4 × (5 × 4) × 1 = 80.

Considering the case that the units digit is 6: The units digit can be filled in 1 way. Having filled this, the digits available for the thousands place are 3, 4, 5, 7 i.e. there are further 4 possibilities. Next, we are left with 5 digits and two of which have to be arranged i.e. can be done in 5 × 4 ways. Thus total possible numbers with unit’s digit being 4 or 6 will be 4 × (5 × 4) × 1 = 80.

Thus the total number of even numbers greater than 3000 that can be formed will be 100 + 80 + 80 = 260

APPLICATION OF PERMUTATION-III

ARRANGEMENT OF PEOPLE IN A ROW

Arranging 4 boys and 4 girls such that 4 girls have to be together:

Considering all the four girls as one unit, we have to arrange 5 units in 5 places and this can be done in 5! ways. Now one of the units is of 4 girls, who can be arranged amongst themselves in 4! ways. Thus the total number of ways of arranging is 5! × 4!.

Arranging 4 boys and 4 girls such that no two girls should stand together:

The four boys can first be arranged in 4! ways. Now there will be 5 places (Three between 4 boys and one at each end) for the 4 girls and they could be arranged in 5 × 4 × 3 × 2 ways. Thus the total number of arrangements possible is 4! × 5 × 4 × 3 × 2.

Arranging 4 boys and 4 girls such that girls and boys have to be alternate:

When girls and boys have to be alternate, it would just be either G B G B G B G B or B G B G B G B G. In each of these ways, there are 4 places for the boys and 4 places for the girls and thus they can be arranged in 4! × 4! in each of these. Thus, the total number of arrangements possible is 2 × 4! × 4!.

APPLICATION OF PERMUTATION-IV

QUESTION-BASED ON FACTORS

Factors of a number N refer to all the numbers which divide N completely. These are also called divisors of a number.

Basic formula related to factors of a number:

These are certain basic formulas pertaining to factors of a number N, such that,

\mathrm{N}=\mathrm{p}^{\mathrm{a}} \mathrm{q}^{\mathrm{b}} \mathrm{r}^{\mathrm{c}}

Where, p, q, and r are prime factors of the number n.

a, b and c are non-negative powers/ exponents

  • Number of factors of N = (a+1)(b+1)(c+1)
  • Sum of factors: ( pa+1-1)( qb+1-1)(rc+1-1)/ (p-1) (q-1) (r-1)

 

Let us take an example:

Find the Number and Sum of the factor of 18.

Factors of 18 are 1,2,3,6,9,18

Number of factors = 6

And their Sum is 39 (=1+2+3+6+9+18)

Using the Formula

18=2^1\times3^2

Number of factors of N = (1+1)(2+1)=6

Sum of factors: ( 22-1)( 33-1)/ (2-1)(3-1)=39

APPLICATION OF PERMUTATION-V

Number of odd and even factors:

Let us understand this concept with an example:

Find the number of Even and Odd factors of 72

Prime factorization of  72=23×32

We know the fact that all the factors of 72 will be in the form of 2a×3b.

For ODD factors, the exponent of 2 i.e., a has to be 0 always. Or, the number of ways using 2 for making the combination is 1, i.e., 20. Also, the number of values that the exponent of 3 i.e., b can take is 3 (0, 1 or 2)

Hence the number of odd factors of 72 = 1x3 = 3.

Extending the logic for EVEN factors, we can say that for a factor to be Even, it must contain 2 at least once.

So, a can take values 3 values (1, 2, or 3, note: 0 is not possible for the number to be even) 

b can take 3 values (0, 1 or 2)

So, the number of values a, and b can take are 3 and 3 respectively.

Therefore, the total number of even factors of 72 = 3x3 = 9.

 

Number of factors which are perfect squares:

Find the number of factors that are perfect squares of 72.

Prime factorization of  72=23×32

If we prime factorize any number which is a perfect square, we would observe that in all cases the exponent of all the prime factors of the number to be even only.

For example, 36 is a perfect square 36=22×32, here we can see that the exponent of both 2 and 3 are even.

Again, any factor 72 will be in the form of 2a×3b. For the factors to be perfect squares, all the values a, and b has to be even only.

Or, the possible values which a can take =  0, 2 i.e. 2 values only. Similarly, b can take 0, 2 i.e. 2 values.

Therefore, the different combinations we can have = 2x2 = 4.

Hence, 72 has 4 factors which are perfect squares.

 

Number of factors which are perfect cube:

Find the number of factors that are perfect cubes of 72.

Prime factorization of  72=23×32

If we prime factorize any number which is a perfect cube, we would observe that in all cases the exponent of all the prime factors of the number to be a multiple of 3.

For example, 27 is a perfect cube 27=33, here we can see that the exponent of 3 is a multiple of three.

Again, any factor 72 will be in the form of 2a×3b. For the factors to be perfect cubes, all the values a, and b has to be divisible by 3.

Or, the possible values which a can take =  0, 3 i.e. 2 values only. Similarly, b can take 0 i.e. 1 value.

Therefore, the different combinations we can have = 2x1 = 2.

Hence, 72 has 2 factors which are perfect cubes.

APPLICATION OF PERMUTATION-VI

DIVISIBILITY OF A FACTOR BY A NUMBER

Let us understand with an example:

Find the number of factors of 72 which are divisible by 8

Prime factorization of  72=23×32

Prime factorization of 8=23

For the factors to be divisible by 8 the factor should be multiple of 8 or 23

Or we can say that the values a should be equal to 3 or greater than 3, and b can take any value.

Hence, the possible values which a can take =  3, i.e. 1 value only. Similarly, b can take 0, 1 and 2 i.e. 3 values.

Therefore, the different combinations we can have = 1x3 = 3.

Hence, 72 has 3 factors which are divisible by 8 i.e. 8, 24, 72.

 

Product of factors:

Let a number be N

It's Prime factorization be N=2ax3bx5c

\large \text {Product of factors}=\text N^{\frac{\text{no. of factors}}{2}}

For example: Find the Product of the factors of 72

\text {Product of factors of 72}=\text 72^{\frac{\text{(3+1)(2+1)}}{2}}

\text {Product of factors of 72}=\text 72^{6}

 

Exponent of Prime P in n!

Where [x] stands for greatest integer value of x\epsilon R.

If m is the index of highest power of a prime p that divides n! then 

m=\left[\frac{n}{p} \right ]+\left[\frac{n}{p^{2}} \right ]+\left[\frac{n}{p^{3}} \right ]+.............

Example: Find the number of trailing zeros in 20!

Solution: 

10 can be written as 2 x 5

If you want to figure out the exact number of zeroes, you would have to check how many times the number N is divisible by 10.

When dividing N by 10, factors of 10 will be the smaller of the power of 2 or 5.

Number of trailing zeros is going to be the power of 2 or 5, whichever is lesser.

m=\left[\frac{20}{2} \right ]+\left[\frac{20}{2^{2}} \right ]+\left[\frac{20}{2^{3}} \right ]+\left[\frac{20}{2^{4}} \right ]+\left[\frac{20}{2^{4}} \right ].............=18

m=\left[\frac{20}{5} \right ]+\left[\frac{20}{5^{2}} \right ]+\left[\frac{20}{5^{3}} \right ]+.............=4

Hence number of trailling zeros are 4

Study it with Videos

APPLICATION OF PERMUTATION-I
APPLICATION OF PERMUTATION-II
APPLICATION OF PERMUTATION-III

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Books

Reference Books

APPLICATION OF PERMUTATION-V

Mathematics Textbook for Class XI

Page No. : 142

Line : 1

APPLICATION OF PERMUTATION-VI

Mathematics Textbook for Class XI

Page No. : 142

Line : 1

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