Total possibilities of forming 5 letter words with alphabets D I G I T is 5*4*3*2*1

Here, there are 5 possibilities for the first position, 4 for the second position, 3 for the third position,
2 for the second position, and 1 for the final position

However, since we have 2 I's, we divide the total possibilities by 2

Total possibilities with digits D I G I T are \(\frac{5*4*3*2*1}{2} = 60\)

Possibilities when the two I's are together
Consider the 2 I's as 1 unit. Let's call it X.
For the digits X D G T, Total possibilities are 4*3*2*1 = 24

Possibilities where I's are not together = Total possibilities - possibilities when they come together = 60-24 = 36

Therefore, the total possibilities in which the two occurrences of I aren't together is 36(Option D)
_________________

You've got what it takes, but it will take everything you've got

hello again pushpitkc, generis, niks18

Here is my solution, its a bit different unlike others` i replaced D,G,I,T, I with following letters \(A\), \(B\), \(C\), \(D\), \(E\), so here, i simply modified question a bit yet arrived at the correct solution.

Using these letters, how many 5-letter strings can be formed in which the two occurrences of the letters A and E are separated by at least one other letter?

So here is my reasoning

\(A\), \(B\), \(C\), \(D\), \(E\)

Step One Fix \(E\), and count number of ways of \(A\) which is 3! * 2 = 12 (we mutiply by 2, because A can be on either left or right side)

Step Two Fix \(A\), and count number of ways of \(E\) which is 3! * 2 = 12 (we mutiply by 2, because E can be on either left or right side)

Step Three Now we have to count number of ways of these three letters \(B\), \(C\), \(D\) which is 3! * 2 = 12 ( mutiply by 2, because ORDER MATTERS/ PERMUTATION)

So total number of ways is \(12+12+12 = 36\)

i doubt that my third step is correct...though ... any ideas ? Quick tips? please

pushpitkc, generis niks18 did you see this post
_________________

In English I speak with a dictionary, and with people I am shy.

Hi Bunuel, can I check why the no of ways when the Is are placed together is 4!? It can be located _D_G_T_. Why isn't it 4 ways? Thanks in advance!

In this stage, why did we have to select 2 spaces and not just place the 1st "I" (4 ways) and the 2nd "I" (3 ways)? For a total of 12 ways?