Combination of Cups | PRMO-2018 | Problem 11

There are several tea cups in the kitchen, some with handle and the others without handles. The
number of ways of selecting two cups without a handle and three with a handle is exactly 1200.
What is the maximum possible number of cups in the kitchen ?

We assume that the number of cups with handle=m and number of cups without handle =n

Can you now finish the problem ..........

The way of select 3 cups with a handle from m cups = \(m_{C_3}\) and select 2 cups without a handle from n cups =\(n_{C_2}\)

Can you finish the problem........

We assume that the number of cups with handle=m and number of cups without handle =n

The way of select 3 cups with a handle from m cups = \(m_{C_3}\) and select 2 cups without a handle from n cups =\(n_{C_2}\)

Therefore \(m_{C_3} \times n_{C_2} \)=1200

\(\Rightarrow \frac {m(m-1)(m-2)}{6} \times \frac{n(n-1)}{2}\)=\(2^4 \times 3 \times 5^2\)

\(m=4,y=25\) and \(m=5,n=16\)

Therefore \(m+n\) be maximum i.e \(m=4,n=25\)

Maximum possibile cups=25+4=29

AMC - AIME Boot Camp... for brilliant students.

Use our exclusive one-on-one plus group class system to prepare for Math Olympiad

Learn More

Subscribe to Cheenta at Youtube