Question

What is the ratio of the number of arrangements of letters of the word ASHOK and GEETA.

Answer 1

First of all, we will find the number of arrangements of the letters in the word ASHOK and the word GEETA using the formulas of permutation. And then divide the number of arrangements to write the required ratio.

Complete step-by-step answer:
We will find the number of arrangements of letters in both the given words, that are ASHOK and GEETA.
In the word ASHOK, there are five letters and none of them is repeating. Each of the letters can be placed in any of the five positions of the word.
Now, we know that if there are $n$ distinct objects that need to be arranged in $n$ distinct places, then there are $n!$ ways of doing it.
Hence, the letters of the word ASHOK can be arranged in 5! Ways.
Next, we will find the number of arrangements of all the letters of the word GEETA
Here, we can observe the letter E is repeating two times.
And if we have $n$ objects and ${n_1}$ is the number of times one object is repeated, ${n_2}$ is the number of times second object is repeated, then the number of arrangements is given by $\dfrac{{n!}}{{{n_1}!{n_2}!...}}$
In the word GEETA , there are a total of 5 words, from which G, T, and A are coming once and E is repeated twice.
Then the number of arrangements is $\dfrac{{5!}}{{2!}}$
Next, we have to find the ratio of the number of arrangements of the letters in the words ASHOK and GEETA.
$\dfrac{{5!}}{{\dfrac{{5!}}{{2!}}}} = \dfrac{{2!}}{1}$
We know that $n! = n.\left( {n - 1} \right)\left( {n - 2} \right).....3.2.1$
Therefore, $\dfrac{{2!}}{1} = \dfrac{2}{1}$
On expressing it in ratio, we get, 2:1

Note: If there are $n$ distinct objects that need to be arranged in $n$ distinct places, then there are $n!$ ways of doing it. If we have $n$ objects and ${n_1}$ is the number of times one object is repeated, ${n_2}$ is the number of times second object is repeated, then the number of arrangements is given by $\dfrac{{n!}}{{{n_1}!{n_2}!...}}$. If the order of the arrangement matters, then we use permutation.