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# Combination of Cups | PRMO-2018 | Problem 11 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 ?

• • • ### Key Concepts

Algebra

combination

maximum

Answer: PRMO-2018, Problem 11

Pre College Mathematics

## Try with Hints

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 = and select 2 cups without a handle from n cups = 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 = and select 2 cups without a handle from n cups = Therefore =1200 =  and Therefore be maximum i.e Maximum possibile cups=25+4=29

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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 ?

• • • ### Key Concepts

Algebra

combination

maximum

Answer: PRMO-2018, Problem 11

Pre College Mathematics

## Try with Hints

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 = and select 2 cups without a handle from n cups = 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 = and select 2 cups without a handle from n cups = Therefore =1200 =  and Therefore be maximum i.e Maximum possibile cups=25+4=29

## Subscribe to Cheenta at Youtube

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