Permutations and combinations!
Level
2
There are several tea cups in the kitchen, some with handles and others without handles. The number of ways of selecting 2 cups without a handle and 3 cups with a handle is exactly 1200. What is the maximum possible number of cups in the kitchen?
27
29
17
5
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1 solution
Call non-handled cups A and cups with handles B . We need ( 2 A ) ( 3 B ) = 1 2 0 0 .
Expand to 2 A ( A − 1 ) ⋅ 6 B ( B − 1 ) ( B − 2 ) = 1 2 0 0 then solve for A
A 2 − A − B ( B − 1 ) ( B − 2 ) 1 4 4 0 0 = 0 then using the quadratic formula
A = 0 . 5 ± 0 . 5 1 + B ( B − 1 ) ( B − 2 ) 5 7 6 0 0 Discard the minus sign solutions as they give negatives.
From here I just made a table to find integer values of ( A , B ) , but you could use some number theory.
The pairs that work are ( 2 5 , 4 ) , ( 1 6 , 5 ) , and ( 5 , 1 0 ) .
The greatest A + B = 2 5 + 4 = 2 9