Find the number of arrangement of the letters of the word independence. In how many arrangements : (a) do all the vowels always occur together (b) do the vowels never occur together (c) do the word begin with i and end with p?
Answers
No. of letters in INDEPENDENCE : 12
repetitions : 4 Es, 2 Ds, 3 Ns.
therefore no. of ways of arranging the letters of the word INDEPENDENCE is 12! / (4!*3!*2!) since there are twelve letters in total and 3 letters are repeated 4 times,3 times and 2 times respectively.
1) WORDS START WITH P
P is fixed for first place. remaining no. of letters is 11 and here also there are 4 Es, 2 Ds, 3 Ns.
therefore , no. of ways of arranging is 11! / (4!*3!*2!).
2) ALL VOWELS OCCUR TOGETHER
vowels present in the INDEPENDENCE are i and e. But e occurs 4 times. so, totally we must consider 5 vowels.remaining no. of letters is 7 whre there are 3 Ns and 2 Ds.
as the 5 vowels are always together , we must consider them as one object. therefore , there are 8 objects totally (7 letters and one set of vowels).
these 8 objects can be arranged in 8! / (3!*2!) * 5! / 4! ways since, in 8 objects , n and d are repeated 3 times and 2 times respectively and we multiply by(5! / 4!) since the vowels can be arranged internally in that many ways.
3) VOWELS NEVER OCCUR TOGETHER
there are only 2 ways of arranging the word INDEPENDENCE. one way is that all vowels occur together and the other way is that vowels never occur together.
therefore, no.of ways in which vowels never occur together will be the total no. of ways of arranging the word INDEPENDENCE minus the no. of ways in which vowels occur together.
that is : (12! / (4!*3!*2!)) - (8! / (3!*2!) * 5! / 4!)
4) WORDS BEGIN WITH I AND END WITH P
I and P are fixed in 2 positions. remaining no. of letters is 10 where there are 4 Es, 2 Ds, 3 Ns.
therefore, total no. of ways is 10! / (4!*3!*2!)