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Combinatorics Math Olympiad PRMO

Combination of Cups | PRMO-2018 | Problem 11

Try this beautiful problem from PRMO, 2018 based on Combination of Cups. You may use sequential hints to solve the problem.

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:$29$

Source
Suggested Reading

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

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