nS= C4 12=495

nE= C28×C14=112

∴P(E)=112495

S = { 1, 2, 3, 4, .....18 } 

=> n(S) = 18

E1 = {2, 4, 6, 8, 10, 12, 14, 16, 18}

=> n(E1) = 9

E2 = {3, 6, 9, 12, 15, 18 }

=> n(E2) = 6

 E3 =E1∩E2={6, 12, 18} 

=> n(E3) = 3

∴E=E1 ∪ E2 = E1+E2-E3 

=> n(E) = 9 + 6 - 3 =12

 where E = { 2, 3, 4, 6, 8, 9, 10, 12, 12, 14, 15, 16, 18 }

∴P(E)=n(E)n(S)=1218=23

Select a number which ocurs on two dice out of six numbers (1, 2, 3, 4, 5, 6). This can be done in C16, ways.

Now select two distinct number out of remaining 5 numbers which can be done in C25 ways. Thus these 4 numbers can be arranged in 4!/2! ways.

So, the number of ways in which two dice show the same face and the remaining two show different faces is 

 C16×C25×4!2!=720

 =>  n(E) = 720

 ∴PE=72064=59

There are 13 spades ( including one king). Besides there are 3 more kings in remaining 3 suits

Thus   n(E) = 13 + 3 = 16

Hence nE¯=52-16=36 

Therefore, PE=3652=913

Let x, y, z be the number of balls received by the three persons, then

x≥5, y≥5, z≥5 and x+y+z=21

Let u≥0, v≥0, w≥0 then

∴ x + y + z =21

⇒ u + 5 + v + 5 + w + 5 = 21

⇒ u + v + w = 6 

∴Total number of solutions = C3-16+3-1=C28=28

A Committee of 5 persons is to be formed from 6 gentlemen and 4 ladies by taking. 

(i) 1 lady out of 4 and 4 gentlemen out of 6 

(ii) 2 ladies out of 4 and 3 gentlemen out of 6 

(iii) 3 ladies out of 4 and 2 gentlemen out of 6 

(iv) 4 ladies out of 4 and 1 gentlemen out of 6 

In case I the number of ways = C14×C46 = 4 x 15 = 60 

In case II the number of ways = C24×C36 = 6 x 20 = 120 

In case III the number of ways = C34×C26 = 4 x 15 = 60

In case IV the number of ways = C44×C16 = 1 x 6 = 6 

Hence, the required number of ways = 60 + 120 + 60 + 6 = 246

A person can be chosen out of 18 people in 18 ways to be seated between Musharraf and Manmohan. Now consider Musharraf, Manmohan, and the third person, sitting between them, as a single personality, we can arrange them in 17! ways but Musharraf and Manmohan can also be arranged in 2 ways. 

Required number of permutations = 18 x (17!) x 2 = 2 x 18!

There are total 9 letters in the word COMMITTEE in which there are 2M's, 2T's, 2E's.

The number of ways in which 9 letters can be arranged = 9!2!×2!×2! = 45360

There are 4 vowels O,I,E,E in the given word. If the four vowels always come together, taking them as one letter we have to arrange 5 + 1 = 6 letters which include 2Ms and 2Ts and this be done in 6!2!×2! = 180 ways.

In which of 180 ways, the 4 vowels O,I,E,E remaining together can be arranged in 4!2! = 12 ways.

The number of ways in which the four vowels always come together = 180 x 12 = 2160.

Hence, the required number of ways in which the four vowels do not come together = 45360 - 2160 = 43200