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

1)In the given word INDEPENDENCE there are 12 letters which are
`3N, 4E, 2D, 1I, 1P, and 1C`

If the words start with `P`, then there are `11` letters to be filled in `11` spaces. Therefore the total number of ways is`\frac{11!}{3!\times 2!\times 4!}=138600`

2. if the vowels EEEEI are to be kept together and should be treated as one unit. Then can be arranged these vowels in `frac{5!}{4!}` ways. This single unit together with 7 letter will count to units, can be arranged in`frac{ 8 !}{ 3 ! × 2 !}`

Therefore the total number of ways` frac{ 5!}{ 4!} × frac{8!}{ 3! × 2!}= 16800`

3. We can calculate number of ways of arrangement with vowels do not come together by formula

`n = `Total arrangement `– `vowels coming together.

`frac{12!}{ 3! × 2! × 4!}– 16800 = 1663200 – 16800 = 1646400`

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