# Find the number of ways to arrange all the letters in the word “MALAYSIA” if the first letter must be a consonant and the last letter must be a vowel.

Find the number of ways to arrange all the letters in the word “MALAYSIA” if the first letter must be a consonant and the last letter must be a vowel.

well here is the way i tried to answer it : we have 4 consonants and 4 vowels so i came up with this formula 16($^6 P_6$)=11520 now i am not sure if this is the correct answer or the correct way to answer the question .

• What is the probability problem about? Right now your question asks how many different ways there are to re-arrange the symbols such that some conditions are satisfied. – Alvin Lepik Aug 8 '17 at 11:00
• this is the full question that's why i couldn't find an answer for it – Nasser Ali adam Aug 8 '17 at 11:03
• Welcome to MathSE. When you pose a question here, it is expected that you include your own thoughts on the problem. What have you tried? Where are you stuck? – N. F. Taussig Aug 8 '17 at 11:03
• 10. Find the number of ways to arrange all the letters in the word “MALAYSIA”, if: (a) all the As must be together. (b) the first letter must be a consonant and the last letter must be a vowel. – Nasser Ali adam Aug 8 '17 at 11:04
• this is the full question i solved the first part but i cant understand the second one – Nasser Ali adam Aug 8 '17 at 11:05

## 2 Answers

we have 8 letters in total (_ _ _ _ _ _ _ _ _)

4 vowels - A,A,A,I

4 consonants - M,L,Y,S

First letter must be consonant so we can pick any letter from M,L,Y,S so we have 4 options - (4 _ _ _ _ _ _ _)

Last letter must be a vowel we can pick A or I so only 2 choices

• choosing A
• choosing I

## Choosing A

we are left with 6 letters 3 consonants and 3 vowels (A,A,I)

so number of possibilities are (6p6)/2!

## Choosing I

we are left with 6 letters 3 consonants and 3 vowels (A,A,A)

so number of possibilities are ($\frac{^6p_6}{3!}$)

so total is 4*(6p6/2! + 6p6/3!) = 1920

Your first question answer should be 6720

There are $4$ ways for a consonant to be in the first position, and $4$ ways for the last letter to be a vowel. Two letters are now taken, so there are $6!$ ways to arrange the remaining 6 letters. This gives $4\cdot4\cdot6!=11520$ permutations as your answer if the three As are not treated as identical. If they are, then there are $3!$ ways to arrange them, so divide to get $\dfrac{11520}{3!}=1920$ as your answer.