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 ?
Algebra
combination
maximum
Answer:$29$
PRMO-2018, Problem 11
Pre College Mathematics
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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