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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 ?

• $24$
• $29$
• $34$

### Key Concepts

Algebra

combination

maximum

Answer:$29$

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 = $m_{C_3}$ 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

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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 ?

• $24$
• $29$
• $34$

### Key Concepts

Algebra

combination

maximum

Answer:$29$

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 = $m_{C_3}$ 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

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