Question

Counting subsets

4. (a). How many ways are there to rearrange (all of) the letters of SPIDERMAN so that all of the consonants appear in alphabetical order (not necessarily consec- utively)? (b). How many ways are there to rearrange (all of) the letters of SPIDERMAN so that the word PRIDE appears together as a word somewhere? (c). How many ways are there to rearrange (all of) the letters of SPIDERMAN that begin and end with a vowel? (d). How many ways are there to rearrange (all of) the letters of SPIDERMAN so that the three vowels appear consecutively?

Answer 1

a)

First arrange the consonants in alphabetical order

SPIDERMAN

DMNPRS

There are 3 positions remaining and we can choose the 3 positions from 9 places in 9C3 ways

So there are

9!/(3!(9−3)!)=84 possible words

b)

In 'SPIDERMAN' rearrangements consider 'PRIDE' as a single element. So it is a rearrangement of 5 elements including 'PRIDE'.

So there are 5!=5*4*3*2*1=120 possible words

c) In 'SPIDERMAN' there are 3 vowels : A,E,I

In 3C2 ways we can select 2 vowels for the 2 positions. In 3C1 ways we can select position 1 and 2C1 ways we can select end position(2C1 because already use one letter). So 3C2*3C1*2C1=3*3*2=18

Now the remaining 7 positions can make 7!=5040 words. So total=18*5040=90720 words.

d) 3 vowels can be arranged in 3!=6 ways and treat it as a single element. Now 7 elements can be arranged in 7! ways

So total=3!*7!=6*5040=30240 words