Answer to Question #98512 in Discrete Mathematics for Ahmed

Question

(a) In how many different ways can the letters of the word wombat be arranged?
(b) In how many different ways can the letters of the word wombat be arranged if the letters wo
must remain together (in this order)?
(c) How many different 3-letter words can be formed from the letters of the word wombat? And
what if w must be the first letter of any such 3-letter word?

Answer 1

(a) wombat consists of 6 distinct letters. Thus, Number of distinct words formed are $6!=720words.$

(b) Number of words in which 'wo' remain together (in the same order) are found by treating 'wo' as a single entitty. Thus we have 5 distinct elements. Thus, such words are $5!=120$ (Answer)

(c) In order to create, 3 letter words we need to first select 3 letters from the 6. This is done in $C(6,3)=20$ ways. Now as all letters are dstinct the chosen letters can be arranged in $3!=6$ ways. Thus, number of such words formed are $=P(6,3)=20*6=120$. (Answer)

Now, if we fix the first letter of these 3 letter words as 'w'. Then the remaining 2 letters can chosen and arranged in $=P(5,2)=5!/2!=60$ ways. Thus, 60 (Answer) 3-letter words can be formed using letters of the word 'wombat' such that first letter is 'w'.

Formulas used:

1. $C(n,r)=n!/[(n-r)!*r!]$ represents number of ways of choosing r objects out of n given objects.
2. $P(n,r)=n!/(n-r)!$ represents number of ways of choosing and arranging r objects out of n given objects.