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

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 ?

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

Algebra

combination

maximum

But try the problem first...

Answer:$29$

Source

Suggested Reading

PRMO-2018, Problem 11

Pre College Mathematics

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

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

Contents

[hide]

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

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 ?

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

Algebra

combination

maximum

But try the problem first...

Answer:$29$

Source

Suggested Reading

PRMO-2018, Problem 11

Pre College Mathematics

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

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

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

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