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Given: We have 6 letters

To Find: Number of ways to arrange letters P,E,N,C,I,L

Condition: N is always next to E

Here we need EN together in all arrangements.

So, we will consider EN as a single letter.

Now, we have 5 letters, i.e. P,C,I,L and ‘EN’.

5 letters can be arranged in5P5ways

⇒5P5

⇒\(\frac{5!}{(5-5)!}\)

⇒\(\frac{5!}{0!}\)

⇒ 120

In 120 ways we can arrange the letters of the word ‘PENCIL’ so that N is always next to E.

Given: We have 12 letters

To Find: Number of words formed with Letter of the word ‘PERMUTATIONS.

The formula used: The number of permutations of n objects, where p1 objects are of one kind, p2 are of the second kind, ..., pk is of a kth kind and the rest if any, are of a different kind is =\(\frac{n!}{p_1!p_2! ....p_k!}\)

In the word ‘PERMUTATIONS’ we have 2 T’s.

We have to start the word with P and end it with S, hence the first and last position is occupied with P and S respectively.

As two positions are occupied the remaining 10 positions are to be filled with 10 letters in which we have 2 T’s.

NOTE:- Unless specified , assume that repetition is not allowed.

Let us represent the arrangement

Hence,

The first place is occupied by P = 1 way

The last place (12th) is occupied by S = 1 way

For the remaining 10 places:

Using the above formula

Where,

n=10

p1=2

\(\frac{10!}{2!}\)= 1814400

Total number of ways 1x1814400x1 = 1814400 are ways.

In 1814400 ways the letters of the word ‘PERMUTATIONS’ can be arranged if each word starts with P and ends with S.

To find:

Number of 4 digit numbers when a digit may be repeated any number of times The first place has possibilities of any of 9 digits.

(0 not included because 0 in starting would make the number a 3 digit number.)

The second place has possibilities of any of 10 digits.

The third place has possibilities of any of 10 digits.

The fourth place has possibilities of any of 10 digits.

Since repetition is allowed.

So there are 9 × 10 × 10 × 10 = 9000 4-digit numbers when a digit may be repeated any number of times.

In forming a 3 digit number the 100’s place can be occupied by any 9 out of 10 digits (0 not included because it will lead to formation of 2 digit number.)

The 10’s place can be occupied by any of the remaining 9 digits (here 0 can or cannot be used.)

In one’s place any of the remain 8 digits can be used.

So total 3-digit numbers with no digit repeated are: 9 × 9 × 8 = 648.

Given: We have 6 letters

To Find: Number of words formed with Letter of the word ‘CHEESE.’

The formula used: The number of permutations of n objects, where p1 objects are of one kind, p2 are of the second kind, ..., pk is of a kth kind and the rest if any, are of a different kind is =\(\frac{n!}{p_1!P_2! ......P_k!}\)

Suppose we have these words – C,H,E1,E2,S,E3

Now if someone makes two words as CHE1E3SE2 and CHE2E3SE1

These two words are different because E1, E2 and E3 are different but we have three similar E’s hence, in our case these arrangements will be a repetition of same words.

In the word CHEESE, 3 E’s are similar

∴ n = 6, p1 = 3

\(\frac{6!}{3!}\)=\(\frac{720}{6}\)= 120

In 120 ways the letters of the word ‘CHEESE’ can be arranged.

To find: number of numbers that can be formed from the digits 1, 3, 5, 9 if repetition of digits is not allowed

Forming a 4 digit number:4!

Forming a 3 digit number: 4C3 × 3!

Forming a 2 digit number: 4C2 × 2!

Forming a 1 digit number:4

So total number of ways = 4! + ( 4C3 × 3!) + (4C2 × 2!) + 4

= 24 + 24 + 12 + 4

= 64

Given: We have 5 digits, i.e. 4,5,6,7,8

To Find: Number of numbers divisible by 5

Condition: (i) Number should be between 4000 and 5000

(ii) Repetition of digits is allowed

Here as the number is lying between 4000 and 5000, we can conclude that the number is of 4-digits and the number must be starting with 4.

Now, for a number to be divisible by 5 must ends with 5

Let us represent the 4-digit number

Therefore,

The first place is occupied by 4 = 1 way

The fourth (last) place is occupied by 5 = 1 way

The second place can be filled by 5 numbers = 5 ways

The third place can be filled by 5 numbers = 5 ways

Total numbers formed = 1 × 5 × 5 × 1 = 25

There are 25 numbers which are divisible by 5 and lying between 4000 and 5000 and can be formed from the digits 4, 5, 6, 7, 8 with repetition of digits.

Given: We have 5 digits i.e. 2,3,4,5,6

To Find: Number of 3-digit numbers

Condition: (i) Number should be greater than 600

(ii) Repetition of digits is allowed

For forming a 3 digit number, we have to fill 3 vacant spaces.

But as the number should be above 600, hence the first place must be occupied with 6 only because no other number is greater than 6.

Let us represent the 3-digit number

So the first place is filled with 6 = 1 ways

Second place can be filled with 5 numbers = 5 ways

Third place can be filled with 5 numbers = 5 ways

Total number of ways = 1 × 5 × 5 = 25

Total number of 3-digit numbers above 600 which can be formed by using the digits 2, 3, 4, 5, 6 with repetition allowed is 25