Question

There are several tea cups in the kitchen,some with handles 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?

Answer 1

Let the no of cups with handle = x

Let the no of cups without handle = y.

Acc to the question, the number of ways of selecting two cups without a handle and three with a handle is exactly 1200.

That means, x(x-1)(x-2)/6 * y(y-1)/2 = 1200.

or, x * (x – 1) * (x – 2) * y * (y – 1) = 14400.

Now, breaking 14400 into five such factors we get,

x * (x – 1) * (x – 2) * y * (y – 1) = 4 * 3 * 2 * 25 * 24

Therefore, x = 4; y = 25.

Not just this but also few other combinations are possible out of this factorisation, like (16 × 15 × 5 × 4 × 3) & ( 10 × 9 × 8 × 5 × 4).

But our target is to get the maximum possible number of cups in the kitchen.

That's why we consider x = 4; y = 25.

Therefore, max no of cups possible = 25 + 4 = 29. [Ans]

Answer 2

Let  the number  of  cups  without  handle x and number  of  cups woith handles  be  y

Number  of  ways of selecting two cups without a handle:-$^{x}C_2$

Number  of  ways  of selecting  3  cups  with handle :- $^{y}C_3$

Total number  of  ways = 1200

$^{x}C_2\times^{y}C_3=1200\\\;\\\frac{x(x-1)}{2}\times\frac{y(y-1)(y-2)}{3!} =1200\\\;\\x(x-1)(y)(y-1)(y-2)=1200\times6\\\;\\x(x-1)(y)(y-1)(y-2)=25\times24\times4\times3\times2\\\;\\x=25\;,y=4\\\;\\or\\\;\\x(x-1)(y)(y-1)(y-2)=16\times15\times5\times4\times3\\\;\\x=16\;,y=5\\\;\\or\\\;\\x(x-1)(y)(y-1)(y-2)=5\times4\times10\times9\times8\\\;\\x=5\;,y=10\\\;\\\text{For maximum value of x+y}\\\;\\x=25\;,y=4\\\;\\\textbf{Number of cups=29}$