Question

Suppose 11 students show up for a meet-the-prof session to which Professor Grumble has brought 43 identical cookies (a) How many ways can the cookies be distributed if everyone except Prof. Grumble is guaranteed at least 1 cookie? (Include solutions in which Prof. Grumble does get one or more cookies, as well as solutions in which he gets 0 cookies.) (b) How many ways can the cookies be distributed if Prof. Grumble gets at least 1 and everyone else gets at least 3?

Answer 1

a)
1X2 111243
with $x_1,x_2\cdots x_{11}\geq 1$

Let

T1y11,2 = y2 + I » » . 1 = Y11+1 T11y11 1

Equation becomes

(y11)(y2+1).(y11 1)x1243 43

So

43- 11 = 32 1y2 .Y1112

The number of solutions for this equation is

53 4212 1V 12 1 11

The number of solutions where Prof. Grumble gets zero cookies corresponds to solutions of

y1y2+ Y11= 32

Which means $\binom{42+11-1}{11-1}=\binom{52}{10}$

The number of solutions where he does get at least one cookie is the difference

53 52 11 10

b) This corresponds to
1X2 111243
where
3, x 12 1, 2

Let $x_k=y_k+3,1\le k\le 11, x_{12}=y_{12}+1$ where $y_1,y_2\cdots y_{12}\geq 0$

Then $(y_1+3)+(y_2+3)\cdots +(y_{11}+3)+y_{12}+1=43$

So that

1y2 y12=43-3 11-1=9

Total number of solutions to this is $\binom{9+12-1}{12-1}=\binom{20}{11}$