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 with I and end in P?

Explanation

(i)In the word INDEPENDENCE there are 12 letters, of which N appears 3 times, E appears 4 times, D appears 2 times and the rest all are different.
When the words start with P, then there are 11 letters to be filled in 11 spaces. Therefore the total number of ways is
`(11!)/(30!×2!×4!) = 138600.`


2. The vowels EEEEI are to be kept together and should be treated as one unit.
Then these vowels can be arranged in `(5!)/(4!)` ways. This single unit together with 7 letter will count to units, can be arranged in `(8!)/(3!×2!)`.
Therefore the total number of ways `(5!)/(4!)×(8!)/(3!×2!) = 16800. `


3. Number of ways of arrangement with vowels do not come together
= Total arrangement – vowels coming together. `
= (12!)/(3!×2!×4!) – 16800 = 1663200 – 16800 = 1646400. `


4. When the words start with I and ends with P, then there are 10 letters to be filled in 10 spaces.
Therefore the total number of ways is `(10!)/(3!×2!×4!) = 12600`