Question

Find the exponent of 10 in ${}^{75}{{C}_{25}}$.

Answer 1

Hint: To find the exponent of 10 in ${}^{75}{{C}_{25}}$, we should learn the expansion of ${}^{n}{{C}_{r}}$, which is equal to $\dfrac{n!}{r!\left( n-r \right)!}$. Also, we should know that, power of some positive prime integer ‘m’ which is $\le n$, in $n!$ is $\left[ \dfrac{n}{m} \right]+\left[ \dfrac{n}{{{m}^{2}}} \right]+\left[ \dfrac{n}{{{m}^{3}}} \right]+\left[ \dfrac{n}{{{m}^{4}}} \right]+.....$, where $\left[ . \right]$ represents the greatest integer number.

Complete step-by-step answer:
We know that ${}^{75}{{C}_{25}}$ can be expressed using the formula ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$, as ${}^{75}{{C}_{25}}=\dfrac{75!}{25!\left( 75-25 \right)!}$, where n = 75 and r = 25.
$\Rightarrow {}^{75}{{C}_{25}}=\dfrac{75!}{25!\left( 50 \right)!}$
Now, in this question, we have to find the exponent of 10 in ${}^{75}{{C}_{25}}$, for that we have to find the exponent of 10 in $75!$, $25!$ and $50!$.
As we know, 10 can be represented as $10=5\times 2$, so the minimum of the exponents of 5 or 2 will be the exponent of 10 in respective numbers.
Now, let us find the exponent of 5 in $75!$, as we know that power of some positive prime integer ‘m’ which is $\le n$, in $n!$ is $\left[ \dfrac{n}{m} \right]+\left[ \dfrac{n}{{{m}^{2}}} \right]+\left[ \dfrac{n}{{{m}^{3}}} \right]+\left[ \dfrac{n}{{{m}^{4}}} \right]+.....$, where $\left[ . \right]$ represents the greatest integer number.
So exponent of 5 is$\left[ \dfrac{75}{5} \right]+\left[ \dfrac{75}{{{5}^{2}}} \right]+\left[ \dfrac{75}{{{5}^{3}}} \right]+.....$
$=\left[ \dfrac{75}{5} \right]+\left[ \dfrac{75}{25} \right]+\left[ \dfrac{75}{125} \right]+.....$
$=\left[ 15 \right]+\left[ 3 \right]+\left[ \dfrac{3}{5} \right]+.....$
$=15+3+0$
$=18$
Now, let us find the exponent of 2 in $75!$,
So, exponent of 2 is$\left[ \dfrac{75}{2} \right]+\left[ \dfrac{75}{{{2}^{2}}} \right]+\left[ \dfrac{75}{{{2}^{3}}} \right]+.....$
$=\left[ \dfrac{75}{2} \right]+\left[ \dfrac{75}{4} \right]+\left[ \dfrac{75}{8} \right]+\left[ \dfrac{75}{16} \right]+\left[ \dfrac{75}{32} \right]+\left[ \dfrac{75}{64} \right]+.....$
$=\left[ 37.5 \right]+\left[ 18.75 \right]+\left[ 9.3 \right]+\left[ 4.6 \right]+\left[ 2.3 \right]+\left[ 1.15 \right]+\left[ 0.57 \right].....$
$=37+18+9+4+2+1+0$
$=71$
As the exponent of 5 is smaller than exponent of 2, that is, 18, we can say that exponent of 10 in $75!$ is 18.
Similarly, we will find the exponent of 10 in $25!$.
Now, let us find the exponent of 5 in $25!$,
So, exponent of 5 is$\left[ \dfrac{25}{5} \right]+\left[ \dfrac{25}{{{5}^{2}}} \right]+\left[ \dfrac{25}{{{5}^{3}}} \right]+.....$
$=\left[ \dfrac{25}{5} \right]+\left[ \dfrac{25}{25} \right]+\left[ \dfrac{25}{125} \right]+.....$
$=\left[ 5 \right]+\left[ 1 \right]+\left[ \dfrac{1}{5} \right]+.....$
$=5+1+0$
$=6$
Now, let us find the exponent of 2 in $25!$,
So, exponent of 2 is$\left[ \dfrac{25}{2} \right]+\left[ \dfrac{25}{{{2}^{2}}} \right]+\left[ \dfrac{25}{{{2}^{3}}} \right]+.....$
$=\left[ \dfrac{25}{2} \right]+\left[ \dfrac{25}{4} \right]+\left[ \dfrac{25}{8} \right]+\left[ \dfrac{25}{16} \right]+\left[ \dfrac{25}{32} \right]+......$
$=\left[ 12.5 \right]+\left[ 6.25 \right]+\left[ 3.12 \right]+\left[ 1.56 \right]+\left[ 0.78 \right]......$
$=12+6+3+1+0$
$=22$
As the exponent of 5 is smaller than exponent of 2, that is, 6, we can say that exponent of 10 in $25!$ is 6.
Similarly, we will find the exponent of 10 in $50!$.
Now, let us find the exponent of 5 in $50!$,
So, exponent of 5 is$\left[ \dfrac{50}{5} \right]+\left[ \dfrac{50}{{{5}^{2}}} \right]+\left[ \dfrac{50}{{{5}^{3}}} \right]+.....$
$=\left[ \dfrac{50}{5} \right]+\left[ \dfrac{50}{25} \right]+\left[ \dfrac{50}{125} \right]+.....$
$=\left[ 10 \right]+\left[ 2 \right]+\left[ \dfrac{2}{5} \right]+.....$
$=10+2+0$
$=12$
Now, let us find the exponent of 2 in $50!$,
So, exponent of 2 is$\left[ \dfrac{50}{2} \right]+\left[ \dfrac{50}{{{2}^{2}}} \right]+\left[ \dfrac{50}{{{2}^{3}}} \right]+.....$
$=\left[ \dfrac{50}{2} \right]+\left[ \dfrac{50}{4} \right]+\left[ \dfrac{50}{8} \right]+\left[ \dfrac{50}{16} \right]+\left[ \dfrac{50}{32} \right]+\left[ \dfrac{50}{64} \right]+......$
$=\left[ 25 \right]+\left[ 12.5 \right]+\left[ 6.25 \right]+\left[ 3.12 \right]+\left[ 1.56 \right]+\left[ 0.78 \right]......$
$=25+12+6+3+1+0$
$=47$
As the exponent of 5 is smaller than the exponent of 2, that is, 12, we can say that exponent of 10 in $50!$ is 12.
Now, to find the exponent of 10 in ${}^{75}{{C}_{25}}$ which is equal to $\dfrac{75!}{25!50!}$, we will put the exponents of respective factorials at their respective places, so we get,
Exponent of 10 in ${}^{75}{{C}_{25}}$ = $\dfrac{{{10}^{18}}}{{{10}^{6}}\times {{10}^{12}}}$
$=\dfrac{{{10}^{18}}}{{{10}^{18}}}$
$={{10}^{0}}$
Hence, the exponent of 10 in ${}^{75}{{C}_{25}}$ is 0.

Note: The possible mistake one can make is while finding the exponent of 10 in $75!$, $25!$ and $50!$, that is, one can find an exponent of 10 without finding an exponent of 5 and 2 which will lead to the wrong solution. Also, one can mistake by writing ${{10}^{0}}$ as 1, which can be confusing. Because ${{10}^{0}}$ means an exponent of 10 in ${}^{75}{{C}_{25}}$ is 0.