Try this beautiful problem from PRMO, 2018 based on Combination of Cups.

## Combination of Cups | PRMO | 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 ?

- $24$
- $29$
- $34$

**Key Concepts**

Algebra

combination

maximum

## Check the Answer

But try the problem first…

Answer:$29$

PRMO-2018, Problem 11

Pre College Mathematics

## Try with Hints

First hint

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

Can you now finish the problem ……….

Second Hint

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……..

Final Step

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

## Other useful links

- https://www.youtube.com/watch?v=0yiR_y8YXc0
- https://www.cheenta.com/trapezium-geometry-prmo-2018-problem-5/

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