Question

Counting subsets

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

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