If Jason is on th ecouncil,this reduces the selction pool to only 13 people,out of which we still need to select 4.

So, 13C4 = 715

The total possible cases would be a 5 card hand with no restrictions :52C5 5

The unwanted cases are:

no queens(out of 48 non-queens cards we get 5) 48C5

only 1 queen(out of 4 queens we get 1,and out of 48 non-queens we get 4) 4C1*48C4

Therefore,52C5-(48C5+4C1*48C4) = 108336

If the queen of spades and the four of diamonds must be in hand,we have 50 cards remaining out of which we are choosing 3.

So, 50C3

There are 13 spades,we must include 2: 13C2

There are 13 diamonds,we must include 3: 13C3

Since we can't have more than 2 spades and 3 diamonds,the remaining 2 cards must be pulled out from the 26 remaining clubs and hearts : 26C2

Therefore,13C2*13C3*26C2 = 7250100

If John must be on the committee,we have 11 students remaining,out of which we choose 4. So,11C4

Required number of ways = 8C5*10C6 =  8C3*10C4 = 11760

Possibilities     Bowlers      Batsmen         Number of ways

                         6               9

         1              4                7              (6C4*9C7)

         2              5                6              6C5*9C6

         3              6                5              6C6*9C5

6C4*9C7 = 15 x 36 = 540

6C5*9C6 = 6 x 84 = 504

 6C6*9C5= 1 x 126 = 126

Total = 1170

Two horses A and B, in a race of 6 horses... A has to finish before B

if A finishes 1... B could be in any of other 5 positions in 5 ways and other horses finish in 4! Ways, so total ways 5*4!

if A finishes 2... B could be in any of the last 4 positions in 4 ways. But the other positions could be filled in 4! ways, so the total ways 4*4!

if A finishes 3rd... B could be in any of last 3 positions in 3 ways, but the other positions could be filled in 4! ways, so total ways 3*4!

if A finishes 4th... B could be in any of last 2 positions in 2 ways, but the other positions could be filled in 4! ways, so total ways... 2 * 4! 

if A finishes 5th .. B has to be 6th and the top 4 positions could be filled in 4! ways..

A cannot finish 6th, since he has to be ahead of B

Therefore total number of ways = 5*4! + 4*4! + 3*4! + 2*4! + 4! = 120 + 96 + 72 + 48 + 24 = 360