Question

The highest power of 18 contained in \[^{50}{{C}_{25}}\text{ is:}\]
(a) 3
(b) 0
(c) 1
(d) 2

Answer 1

Hint: First expand the combination terms into factorial. Now, do prime factorization for 18. Find the number of primes which are the factors of 18. By these try to form 18 because you know the number of 18’s in the expression, that answer will be its power.

Complete step-by-step answer:
Given the expression of combination in the question can be written as:
\[^{50}{{C}_{25}}\]
Combinations: It is a mathematical technique that determines the number of possible arrangements in a collection of items where the order of selection does not matter in the combination. You can select the items in any order. The formula is given by:
\[^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\]

By expanding this term using combination definition, we get,
\[^{50}{{C}_{25}}=\dfrac{50!}{25!\left( 50-25 \right)!}\]
By simplifying, we can write the expression as in the form
\[^{50}{{C}_{25}}=\dfrac{50!}{25!25!}\]
Prime factorization of 18: By the definition, we do it as,
By dividing 18 with 3, we get the expression as:
\[18=3\times 6\]
By dividing 6 with 3, we get the expression as:
\[18=3\times 3\times 2\]
By dividing 2 with 2, we get the expression as:
\[18=3\times 3\times 2\times 1\]
We have to stop here, as we have got 1 as the quotient.
\[18={{3}^{2}}\times 2\]
We need to find the powers of 2 and 3 in the given expression. The power of p in n! can be written as:
\[{{E}_{p}}\left( n \right)=\left[ \dfrac{p}{n} \right]+\left[ \dfrac{p}{{{n}^{2}}} \right]+.....+\left[ \dfrac{p}{{{n}^{k}}} \right]\text{ where }{{n}^{k}}>p\]
By the above formula, we can calculate the following values. The power of 2 in factorial of 50, can be written as:
\[=\left[ \dfrac{50}{2} \right]+\left[ \dfrac{50}{{{2}^{2}}} \right]+\left[ \dfrac{50}{{{2}^{3}}} \right]+\left[ \dfrac{50}{{{2}^{4}}} \right]+\left[ \dfrac{50}{{{2}^{5}}} \right]+\left[ \dfrac{50}{{{2}^{6}}} \right]\]
By simplifying, we get it in the form of:
= 25 + 12 + 6 + 3 + 1 + 0
By solving the above sum, we get its value as 47. The power of 2 in the factorial of 25 can be written as
\[\Rightarrow \left[ \dfrac{25}{2} \right]+\left[ \dfrac{25}{{{2}^{2}}} \right]+\left[ \dfrac{25}{{{2}^{2}}} \right]+\left[ \dfrac{25}{{{2}^{4}}} \right]+\left[ \dfrac{25}{25} \right]=12+6+3+1+0\]
We can write the sum as 22. So, the power of 2 in \[\dfrac{50!}{25!25!}\] is (Power of 2 in 50!) – 2 (Power of 2 in 25!)
As in division, both the powers will be subtracted = 47 – 2(22)
So, the power of 2 in the expression = 3…..(i)
The power of 3 in factorial of 50 can be written as:
\[\Rightarrow \left[ \dfrac{50}{3} \right]+\left[ \dfrac{50}{{{3}^{2}}} \right]+\left[ \dfrac{50}{{{3}^{3}}} \right]+\left[ \dfrac{50}{{{3}^{4}}} \right]=16+5+1+0=22\]
The power of 3 in the factorial of 25 can be written as:
\[\Rightarrow \left[ \dfrac{25}{3} \right]+\left[ \dfrac{25}{{{3}^{2}}} \right]+\left[ \dfrac{25}{{{3}^{3}}} \right]=8+2+0=10\]
The power of 3 in the expression can be written (a concept used in 2)
= (Power of 3 in 50!) – 2 (Power of 3 in 25!) = 22 – 2 (10) = 2…..(ii)
By these values of the power of 3 and 2, we can write
\[{{3}^{2}}\times {{2}^{3}}=18\times {{2}^{2}}\]
So, we can form only one 18. So, the highest power of 18 in \[^{50}{{C}_{25}}\] is 1.
Hence, the option (c) is the right answer.

Note: You can also apply directly to 18 itself. But as you know 18 can be formed from the factor of 2 from other numbers. So, we must do prime factorization to find the powers of those only to get the exact value of the powers. If you have more powers of 3, then you can form more numbers of 18’’s/ But here we have only 2.