10C5 x 28C1 x 10C6 = 7266

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

The toys are different; The boxes are identical 

If none of the boxes is to remain empty, then we can pack the toys in one of the following ways 

a. 2, 2, 1 

b. 3, 1, 1 

Case a. Number of ways of achieving the first option 2 - 2 - 1 

Two toys out of the 5 can be selected in 5C2 ways. Another 2 out of the remaining 3 can be selected in 3C2 ways and the last toy can be selected in 1C1 way. 

However, as the boxes are identical, the two different ways of selecting which box holds the first two toys and which one holds the second set of two toys will look the same. Hence, we need to divide the result by 2 

Therefore, total number of ways of achieving the 2 - 2 - 1 option is ways 5C2*3C2= 15 ways

Case b. Number of ways of achieving the second option 3 - 1 - 1

Three toys out of the 5 can be selected in 5C3 ways. As the boxes are identical, the remaining two toys can go into the two identical looking boxes in only one way.

Therefore, total number of ways of getting the 3 - 1 - 1 option is 5C3 = 10 = 10 ways.

Total ways in which the 5 toys can be packed in 3 identical boxes

= number of ways of achieving Case a + number of ways of achieving Case b= 15 + 10 = 25 ways.