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What is the highest power of 18 contained in $^{50} C_{25}$ ?

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answer is C.

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Detailed Solution

Given tern is

^{50} C_{25}

.
Now the formula of combination can be given as,

^{n} C_{r} = \frac{n!}{r (n! - r!)}

Thus, the given term can be written as,

\begin{matrix} ^{50} C_{25} = \frac{50!}{25! (50! - 25!)} \end{matrix}

^{50} C_{25} = \frac{50!}{(25!) (25!)}

Now, the factors of 18,

18 = 3 \times 2 \times 2

18 = 3 \times 2^{2}

Here, 2′s and 3′s in 50! and 25! Need to be find out.
2′s in 50!

\begin{matrix} = [\frac{50}{2}] + [\frac{50}{4}] + [\frac{50}{8}] + [\frac{50}{16}] + [\frac{50}{32}] \\ = 25 + 12 + 6 + 3 + 1 \\ = 47 \\ 3^{'} \sin 50! \\ = [\frac{50}{3}] + [\frac{50}{9}] + [\frac{50}{27}] \\ = 16 + 5 + 1 \\ = 22 \end{matrix}

Similarly, no. of 2s in

25! = 22

and no. of 3s in

50! = 10

Therefore In \frac{50!}{(25!) (25!)} \to

no. of

2^{'} s = 47 - 44 = 3

and no. of

3^{'} s = 22 - 20 = 2

.
So, only

18^{1}

is possible.
Here, highest power of 18 is 1.
Hence the answer is 3.

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