0

How many arrangements can be made using all the letters of the word DIAGONISE such that each of these starts with a vowel and ends with a consonant?

Kme
  • 11
  • How many ways can you choose a vowel and a consonant? Once this is done, in how many ways can you permute the rest of the letters? – Sarvesh Ravichandran Iyer Oct 20 '16 at 09:23
  • I am confused please provide some solution – Kme Oct 20 '16 at 09:26
  • Since "diagonise" is not actually a word, I suggest you double-check that you have not copied the exercise incorrectly. –  Oct 20 '16 at 09:26
  • What is the problem you are facing ? On this site, you need to explain what you have already tried, else the question is likely to get closed. – true blue anil Oct 20 '16 at 09:27
  • yeah.since it is an exercise so it is correct – Kme Oct 20 '16 at 09:28
  • at the starting place I can use {I,I,A,O,E} so there would be 5 ways to do this. Then at the end I can use 4 of the consonants so there would be 4 ways.Is it correct? – Kme Oct 20 '16 at 09:31
  • @Kme notice that if you start with one I as opposed to the other I -- you still have the same combination. We need to eliminate this over counting ! – Hugh Entwistle Oct 20 '16 at 09:33
  • yeah due to these two I's I got confused – Kme Oct 20 '16 at 09:41

2 Answers2

0

Apologies, my previous answer was incorrect...

Basically, since I appears twice -- you need to separate it into two case statements:

CASE 1: (I is picked as the first vowel)

This will leave you with 4 ways to pick the last consonant, and then $7!$ combinations, (since the elements that we are permutating are now unique! So there is no need to divide by 2 here.

CASE 2: (I is not picked as the first vowel)

In other words, there are only 3 vowels to choose from (since we have already handled the others in Case 1 ). There are still 4 ways to pick the last consonant, and then $\frac{7!}{2}$ combinations (since you now have two identical elements in your 'body of the word')

Add up both of these cases:

Hugh Entwistle
  • 1,239
  • thanks a lot.now I can understand what is happening.Although I tried to understand your last answer but the more I was trying to understand , the more I was thinking whether it is correct or not.This solution gives me the perfect idea about the things without any confusions. Thanks again. – Kme Oct 20 '16 at 11:13
  • Apologies for the first answer being incorrect! Perms and combs is always very involved inspection of the question, I overlooked the question as being simpler than it was.. Glad you understand now! – Hugh Entwistle Oct 20 '16 at 11:25
0

No of vowels = 2I + A + O + E= 5ways

No of Consonants = D + G + N + E =4ways

Remaining words can be arrange in $7!$

so,,result will be

$$\frac{(5)(4)(7!)}{(2!)}$$

avk avk
  • 1