As no two boys are to be together, we need the presence of a girl between every two boys.

The seven girls can be arranged in 7! Ways.

_G_G_G_G_G_G_G_

The five boys can be arranged in 8 available gaps in ⁸P₅ ways.

∴ the total number of arrangements = 7! × ⁸p₅

A million is 1000000(first seven digit’s number). So, we need to find how many numbers of less than 7 digits can be formed using the digits 0, 7 and 8.

This can be done as there are 2 possibilities at this first digit ⇒ 7 and 8

At all other places, any of the three numbers can be present. So there are three possibilities for all other digit’s places, therefore possible numbers

Number of 1 digit numbers = 2

Number of 2 digit numbers = 2 × 3 = 6

Number of 3 digit numbers = 2 × 3 × 3 = 18

Number of 4 digit numbers = 2 × 3 × 3 × 3 = 54

Number of 5 digit numbers = 2 × 3 × 3 × 3 × 3 = 162

Number of 6 digit numbers = 2 × 3 × 3 × 3 × 3 × 3 = 486

Total number of such numbers = 2 + 6 + 18 + 54 + 162 + 486 = 728

First of all we will make circular permutations for all the flowers of a particular colour.

We know that, for permutation of n objects in a circle is = (n - 1)!

So, for 10 flowers of same colour, the permutations in 10 places = (10 - 1)! = 9!

Now, for permutation of the remaining ten flowers of same colour in between the previously placed flowers = 10!

Now, since this is a necklace hence the number of permutations are divided by 2.

Hence, the required answer = (9! × 10!)/2 = 5(9!)²

When L and T are fixed as first and last letters of the word, then we have only 6 letters to arrange.

Hence, required number of words  \(= {{!6} \over {!2 \times !2 \times !2}} = {{720} \over 8} = 90\)\({!2 × !2 × !2 ⇒ because M, A I occurs two times}.\)

Total number of letters in ARRAGEET = 8

In the given word there are 2 “R”, 2 “A” and 2 “E”.

As we know that number of ways of arranging a letters out of which r are of same type, m are of same type, n are of same type and p are of same type = \(\frac{{{\text{a}}!}}{{{\text{r}}!{\text{m}}!{\text{n}}!{\text{p}}!}}\)

So total number of ways = \(\frac{{8!}}{{2!2!2!}} = 5040\)

In a simultaneous throw of two dice –

Number of possible outcomes n(s) = 6 × 6 = 36

Number of favourable events = n(E) = {(1, 6); (2, 5), (3, 4), (4, 3), (5, 2), (6,1)}

⇒ n(E) = 6

∴ Required probability of getting a total of 7 –

\(P\left( E \right) = \frac{{n\left( E \right)}}{{n\left( S \right)}}\)

\(= \frac{6}{{36}} = \frac{1}{6}\)

The first letter can be chosen in 26 ways. Similarly, second letter can be chosen in 26 ways.

Hence, the first 2 letters can be chosen in 26 × 26 ways.

Also, the third letter can be chosen in 26 ways.

Hence, the first 3 letters can be chosen in 26 × 26 × 26 ways.

But the fourth letter should be similar to second letter. This can be done in just 1 way.

Similarly, fifth letter should be similar to first letter. This can be done only in 1 way.

Hence, number of words possible = 26 × 26 × 26 × 1 × 1

= 17576

Case – I         

Committee contains 2 women and 1 men.

∴ No. of ways of forming this committee = (number of ways of choosing 2 women out of 3) × (number of ways of choosing 1 man out of 5)

= ³C₂ ⋅ ⁵C₁ =3 × 5 = 15

Case – II        

Committee contain 3 women and 0 men

No. of ways of forming this committee = number of ways of choosing 3 women out of 3.

= ³C₃  =1

∴ Total number of ways = 15 + 1 = 16

First digit can be 1, 2, 3 only. (four digit number)

Hence, 3 × 3 × 2 × 1 = 18 ways (with no digit repetition)

We know that:

Formula:

\({}_{}^n{P_r} = \frac{{!n}}{{!\left( {n - r} \right)}}\)

∴⁶P₂ × ⁵P₃ =  \(\frac{{!6}}{{!\left( {6 - 2} \right)}} \times \frac{{!5}}{{!\left( {5 - 3} \right)}}\)

⇒⁶P₂ × ⁵P₃ \(= \frac{{!6}}{{!4}} \times \frac{{!5}}{{!2}}\)

⇒⁶P₂ × ⁵P₃ = \(\frac{{6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{4 \times 3 \times 2 \times 1}} \times \frac{{5 \times 4 \times 3 \times 2 \times 1}}{2}\)

⇒⁶P₂ × ⁵P₃ = 30 × 60

⇒⁶P₂ × ⁵P₃  = 1800

x=√17956=134