Question

how many arrangements can be made by the letters of the word DEFINITION if the letter I do not occupy the first or last place?

Answer 1

We are making a 10-letter word. We are told that all the occurrences of the letter I must be on the "inside." There are therefore (83) ways to choose where the I's will go. For every decision about where the I's go, there are (72) ways to decide where the N's go. For every decision about the I's and N's, there are 5! ways to arrange the remaining letters in the 5 remaining vacant places, for a total of
(83)(72)5!words.

b) The I's must be together, so let's tie them together into a single "megaletter" like this: III¯¯¯¯. We are forming 8-letter words from the "letters" D, E, F, III¯¯¯¯, N, N, O, and T.

The location of the two N's can be chosen in (82) ways. For each choice of location for the N's, there are 6! ways to arrange the remaining "letters" in the 6 remaining open slots, giving a total of
(82)6!words.

Answer 2

hmm..i dont think that is right..

Answer 3

ok

Answer 4

if u say so

Answer 5

:/

Answer 6

there r three I's...and these 3 I's can be arranged within 8 places by 8 P 3 / 3! ways

Answer 7

now u have remaining 7 vacant places...which can be filled by 7! /2! ways.

Answer 8

SO TOTAL NO OF ARRANGEMENTS= (8 p 3)* 3! 7! / ( 2! 3!)

Answer 9

isnt it 8P3 / 3! * (7!/2!) ?

Answer 10

yes

Answer 11

ok. thanks.

Answer 12

:)