Let S be the sample space

n(S) = 52C₃ = [( 52 x 51 x 50) / (3 x 2 x 1)] = 13600 / 6 = 22100

Let E = event of getting 3 kings out of 4.

n(E) = 4C₃ = [(4 x 3 x 2 x 1) / (3 x 2 x 1)] = 24/6 = 4

therefore, p(E) = [n(E) / n(S)]

==> p(E) = [ 4 / 22100]

==> p(E) = 1/5525

25! / 23! = 23! X 24 x 25 / ( 23!) = 24 x 25 = 600

(16 – 2)! × 2 = 14! × 2

Since each desired number is divisible by 5, so we must have 5 at the unit place. So, there is 1 way of doing it.

The tens place can now be filled by any of the remaining 5 digits (2, 3, 6, 7, 9). So, there are 5 ways of filling the tens place.

The hundreds place can now be filled by any of the remaining 4 digits. So, there are 4 ways of filling it.

Required number of numbers = (1 x 5 x 4) = 20.

Total 20 persons, 3 always seat together, 17 + 1 =18 delegates can be seated in (18 -1)! Ways = 17!
And now that three can be arranged in 3! Ways.
So, 17! 3! is the correct answer.

Maximum number of such different groups = ABC, ABD,ABE, BCE,BDE,CEA,DEA =7.

n the word, “MATERIAL” there are three vowels A, I, E.

The word ‘OPTICAL’ contains 7 different letters.

When the vowels OIA are always together, they can be supposed to form one letter.

Then, we have to arrange the letters PTCL (OIA).

Now, 5 letters can be arranged in 5! = 120 ways.

The vowels (OIA) can be arranged among themselves in 3! = 6 ways.

 Required number of ways = (120 x 6) = 720.

Number of letters in SMALL is 5, but there are 2 L?s Answer is 5! / 2! = 120 / 2 = 60

Six members can be selected from ten members in