Question

How many arrangements can be made by the letters of the word 'definition' if the letters 'i' do not occupy either the first or the last place?

Answer 1

since there are 10 letters,there are 10 places to fill,but 1st place cannot be filled with i,so for 1st place there are only 8 possibilities....then 2nd place can be fillled with 9 other letters,3rd place,8 other letters,4th place 7 other letters...and so on to give number of arrangements as 8*9*8*7*6*5*4*3*2*1.....but this includes 1 case with i in the end,so overall no. of arrangements is 1 less,that is (8*9*8*7*6*5*4*3*2*1-1)....got it?

Answer 2

that's what i did sort of but the answer is 141 120 and hm, does that sound like theyve got the wrong answer printed?

Answer 3

we seem to be correct....but lets see,if some more talented one attempts this and says otherwise...

Answer 4

lol sounds good, thanks but!

Answer 5

check-
i is repeating 3 tmes, and n is repeating 2 times

Answer 6

we need to form 3 cases and add them :
Case I : first and last places dont have "n"
Case II : first and last places both are fixed with "n"
Case II : first and last places have one "n"

Answer 7

Case I : \((5*4) * \frac{8!}{3!2!}\)

Answer 8

oops,did not notice i repeats 3 times!! but where does n come from??

Answer 9

n is there in the word hehe

Answer 10

no,i m saying constraint is only on i... why consider n

Answer 11

its not a constraint from question, its part of solution.
we need to consider where n goes, to the middle (or) to the first and last.

Answer 12

but why?

Answer 13

yes ganehie,why is n that much important?

Answer 14

becoz n is repeating.
if n is in middle, we have 5 unique characters to pick for first & last
if n in NOT in middle, we have only 4 uniq characters to pick for first & last

makes sense eh ?

Answer 15

we must account for all 3 cases listed

Answer 16

now why the hell that did not strike me!!its little bit above my level....i'll leave it to u ganeshie..

Answer 17

oh its okay... not high level. just we need to patiently work out all 3 cases. lot of arithmetic. painful job... :)

Answer 18

let move to case II

Answer 19

Case II : first and last places both are fixed with "n"

Since, first and last are fixed with n, we have 8 characters to pick for middle.
i - > repeats 3 times

\(\frac{8!}{3!}\)

Answer 20

yesum , i think i get it, if you dont wanna write anymore, its okay, i think i understand :) but if you want, thats cool haha

Answer 21

Case III : first and last places have one "n"
\((2*5) *\frac{8!}{3!}\)

Answer 22

ok nice :)

Answer 23

http://www.wolframalpha.com/input/?i=%285*4%29%288%21%2F%283%212%21%29%29+%2B+8%21%2F3%21+%2B+%282*5%29%288%21%2F3%21%29

Answer 24

you are the god of probability, thank you so much :)

Answer 25

ohhh thank you :D

Answer 26

ohkay wait, so why are we basing all these calculations around n? what are we trying to find that way? how does that tell us if 'i' is not first or last?

Answer 27

i is not first or last - its the given constraint. while working out 3 cases, we accounted for it. let me explain how we accounted for "i" in case I

Answer 28

Case I : first and last places dont have "n"

we have 10 letters in "definition"
i -> 3
n -> 2
remaining 5 -> unique

right ?