Question

Timothy is re-arranging his marble collection. He has five identical blue marbles,
five identical green marbles and three identical black marbles. He can fit exactly five
marbles into a case and must have at least one of each. How many different ways
can he arrange the case in​

Answer 1

Answer:

150

Step-by-step explanation:

If order isn't important then it is the same count as natural solutions to:

b + g + k = 5

with b>0, g>0, 0<k<4 which form a bijection to the solutions to:

b + g + k = 2

b>=0, g>=0, k>=0

there are C(4,2) = 6 such solution.

We could list them (b,g,k):

(1,1,3)

(1,3,1)

(3,1,1)

(1,2,2)

(2,1,2)

(2,2,1)

If order does matter then we simply add together the permutations of each solution set above e.g. (1,1,3) has the count 5!/(1!1!3!) and (1,2,2) has 5!/(1!2!2) permutations.

3*5!/(1!1!3!) + 3*5!/(1!2!2!) = 150

we can check this using an inclusion exclusion method where we first subtract from 3^5 the 3*2^5 ways of getting 0 blue, green or black then add back the 3*1^5 ways of getting 0 of any 2 colours:

3^5 - 3*2^5 + 3*1^5 = 150