Question

How many different words can be formed from the letters of the word GANESHPURI when the vowels always occupy even places (i.e. ${2^{nd}}$, ${4^{th}}$ etc.)?

Answer 1

We can find the number of letters and number of vowels and consonants by counting. Then we can find the number of even places and find the number of ways of arranging the vowels in them by taking the permutations. Then we can find the number of ways of arranging the consonants in the remaining positions by taking the permutations. Then we can take the total number of words by taking their product.

Complete step-by-step answer:
We have the word GANESHPURI.
By counting there are 10 letters, out of which 4 are vowels. The vowels are A, E, U, I.
We need to find the number of words that can be formed when the vowels always occupy the even places.
As there are 10 letters, there will be 10 places out which half will be even places. So, there are 5 even places. We have 4 vowels and need to arrange in the 5 even places. This can be done in ${}^5{P_4}$ways.
We know that ${}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}$. So, we get,
$\Rightarrow {}^5{P_4} = \dfrac{{5!}}{{\left( {5 - 4} \right)!}}$
On simplification, we get,
$\Rightarrow$ ${}^5{P_4} = \dfrac{{5!}}{{1!}}$
We know that $1! = 1$. So, we get,
$\Rightarrow {}^5{P_4} = 5!$
On expanding the factorials, we get,
$\Rightarrow {}^5{P_4} = 5 \times 4 \times 3 \times 2 \times 1$
On taking the product, we get,
$\Rightarrow {}^5{P_4} = 120$
So, the vowels can be arranged in 120 ways.
Now we have 6 consonants left and 6 places left. So, we can arrange the consonants in ${}^6{P_6}$ ways.
We know that ${}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}$. So, we get,
$\Rightarrow {}^6{P_6} = \dfrac{{6!}}{{\left( {6 - 6} \right)!}}$
On simplification, we get,
$\Rightarrow {}^6{P_6} = \dfrac{{6!}}{{0!}}$
We know that $0! = 1$. So, we get,
$\Rightarrow {}^6{P_6} = 6!$
On expanding the factorials, we get,
$\Rightarrow {}^6{P_6} = 6 \times 5 \times 4 \times 3 \times 2 \times 1$
On taking the product, we get,
$\Rightarrow {}^6{P_6} = 720$.
So, we can arrange the consonants in 720 ways.
The number words that can be formed in the given condition is given by the product of the number of ways of arranging the vowels and the consonants.
$\Rightarrow n = 720 \times 120$
On simplifying, we get,
$\Rightarrow n = 86400$
Therefore, the number of words that can be formed is 86400.

Note: We must know that the vowels are the letters A, E, I, O and U. We must note that the condition is given that the vowels are only placed in even places. This doesn’t mean that the consonants cannot take the even places. We must arrange the vowels first and the remaining place is filled by the consonants. We must know that $0! = 1$ and not equal to zero. We must take only the product, not the sum to find the required number of words.