How many can the donuts be distributed among the staff? (1) A box of 13 donuts contains 5 catsup, 2 chocolate, 3 mayo, 3 plain. There are 12 members of the ics department each to have one donut. (a) How many ways can the donuts be distributed among the staff, if they all get one donut? (b) What is Prof. Duncan insists on having a chocolate donut?
(2) (a) How many ways can 10 people be paired up for one on one basketball?
(b) How many ways can 10 people be divided into two teams to play each other?
(1) (a) using the formula $\binom{n+k-1}{k-1}$ I get
16C11 * 13C11 * 14C11 * 14C11
(b) I'm not sure but I think it is
13C2 * 15C10 * 13C10 * 13C10
(2) (a) $\frac{10C2 \cdot 8C2 \cdot 6C2 \cdot 4C2 \cdot 2C2}{5!}$
(b) If the teams are equal then $\frac{10C5 \cdot 5C5}{2!}$
I'm not sure about the case then the teams are not balanced.
 A: In b) part of 2nd question. There is nothing mentioned about balanced teams. So you have to consider other cases also. 
A: Catsup and mayo donuts? Yuch.
For part a, you must leave one donut undistributed, either catsup, chocolate, mayo or plain giving 4 cases.
For the case where you leave out the catsup donut you get the following number of ways:
$\binom {12} 4*\binom 8 2*\binom 6 3*\binom 3 3$ 
Leaving out a chocolate donut:
$\binom {12} 5*\binom 7 1*\binom 6 3*\binom 3 3$ ways
Leaving out a mayo donut:
$\binom {12} 5*\binom 7 2*\binom 5 2*\binom 3 3$ ways
and the plain:
$\binom {12} 5*\binom 7 2*\binom 5 3*\binom 2 2$ ways
Sum those possible ways.
For part b, take out one chocolate donut and distribute the remaining 12 donuts amongst the remaining 11 faculty members in the same manner as above.
Edit: To see the reason, take the first case where we leave out one catsup donut. Now we have 4 catsup, 2 chocolate, 3 mayo, 3 plain. Now we distribute them. First choose 4 of the 12 staff members and give each a catsup donut. Next choose 2 of the remaining 8 staff members to give chocolate donuts, chose 3 of the 6 remaining members to give mayo donuts, and choose the rest, 3 of 3 to give plain. You could distribute in any order, but you'll get the same number (i.e. you could distribute the plain first, etc.).
A: The easiest way to do question 1a is to say that it is the number of different ways to arrange the donuts in a line. Then each member of the department in order comes and takes the next donut, and the one that's left at the end goes back in the box. So it is just $\frac{13!}{5!2!3!3!}$. 
For part 2a, first we decide who the oldest player is paired with. There are $9$ choices. Now we have $8$ players left, so we choose who the oldest remaining one is paired with; there are $7$ choices. Continuing in this manner we get $9\times 7\times 5\times 3\times 1$. (Your answer gives the same number.)
For part 2b, I am pretty sure equal teams is what's intended (since it's talking about basketball, and there are 5 players on a basketball team). So your answer is correct.
A: *

*a)


Since 13 donuts need to be distributed among 12 members, 1 donut has to be left out.
We can leave each type of donut out and distribute remaining 12 and add all the cases.
Leaving catsup donut out: 12!/(4!*2!*3!*3!) 
Leaving chocolate donut out: 12!/(5!*1!*3!*3!)
Leaving mayo donut out: 12!/(5!*2!*2!*3!)
Leaving plain donut out: 12!/(5!*2!*3!*2!)
We can add them all.
b) Here we can always give the chocolate donut to Prof. Duncan. Then the problem becomes: 'A box of 12 donuts contains 5 catsup, 1 chocolate, 3 mayo, 3 plain. There are 11 members of the ics department each to have one donut.'
Repeat part (a) here.
