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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

Check the Answer


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

Check the Answer


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

Subscribe to Cheenta at Youtube


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