Find the number of arrangements of the letters of the INDEPENDENCE. In how many of these arrangements, (i) do the words start with P (ii) do all the vowels always occur together (iii) do the vowels never occur together (iv) do the words begin wit

Explanation

I. The given word contains 12 letters out of which N occurs 3 times, E occurs 4 times, D occurs 2 times and the rest are all different.
Hence, the required number of arrangements
`=(12!)/((3!)xx(4!)xx(2!))=1663200.`
II. (i) After fixing P at the extreme left position, there are 11 letters consisting of 3 N's, 4E's, 2D's, 11 and 1 C.
So, the number of words beginning with P
= number of arrangements of 11 letters out of which there are 3N's, 4E's, 2D's, 11 and 1C
`=(11!)/((3!)xx(4!)xx(2!))=138600.`
(ii) There are 5 vowels in the given word, namely 4E's and 11.
Let us treat them as a single letter EEEEI.
This letter with 7 remaining letters will give us 8 letters in which there are 3 N's, 2 D's, 1 P, 1 C and 1 letter EEEEI.
Number of all such arrangements `=(8!)/((3!)xx(2!))=3360.`
Number of arrangements of 5 vowels, namely 4E' s and 11
`=(5!)/(4!)=5`.
Hence, the number of words in which vowels are together
`=(3360xx5)=16800.`
(iii) Number of arrangements in which vowels do not occur together
`= (total number of arrangements) - (number of arrangements in which vowels occur together) ltbgt `=(1663200-16800) =1646400.`
(iv) Let us fix I at the left end and P at the right end of each arrangement.
Then, we are left with 10 letters, out of which there are 3 N's, 4 E's, 2D's and 1 C.
Hence, the number of words which begin with I and end in `P=(10!)/((3!)xx(4!)xx(2!))=12600.`