The letters D, G, I, I , and T can be used to form 5-letter strings as

The letters D, G, I, I , and T can be used to form 5-letter strings as DIGIT or DGIIT. Using these letters, how many 5-letter strings can be formed in which the two occurrences of the letter I are separated by at least one other letter?

A) 12
B) 18
C) 24
D) 36
E) 48

An easy way to layout this problem:

All the possible DIFFERENT combinations - the impermissible combinations = the total permissible combinations.

For instance, to find all the different combinations, create a fraction where the numerator is the factorial number of digits (5!) and the denominator is factorial the number of repeats (2!). We divide this by the factorial of the number of repeats because there are two Is, let's say we have I1 and I2. If two possible codes are I1I2DGT or I2I1DGT, it doesn't matter which order the I's are in, the code is still IIDGT. Thus, we must divide by this factorial because if we don't, we would have double the actual amount of codes.

So now we have all of the different combinations as (5!/2!), which equations to (5*4*3*2!/2!), which gives us 60 total combinations.

Now, we need to find the amount of impermissible combinations, because two I'ss cant be next to each other. Think of the two I's as one "block": it doesn't matter which I comes first, they just can't be next to eachother. So we have 4 different combinations of impermisslbe numbers:

II D G T
D II G T
D G II T
D G T II

We've already accounted for the different combinations that the D, G, and T can be moved around to form different codes in the first (ALL) part of the equation. We are simply looking for how many ways the II block can be next to each other, so we can subtract it from the ALL part of the equation. This gives us 4! ways that the blocks can be next to eachother, which equals (4*3*2*1) = 24.

Now, we need to take the total amount of combinations (60) and subtract the impermissible combinations where the two Is are next to each other (24).

So our answer is 60 - 24 = 36.

Hope this helps.

Hi,
Could you please explain why the below method is incorrect as I could not find by mistake

- If the 2 I's are separated by 1 letter
3! * 4C2 because there are 4 gaps between D, G and I = 36

- If the 2 I's are separated by 2 letters
3! * 3C2 because now there are 3 gaps. = 18

- If the 2 I's are separated by 3 letters
3! = 6

Total = 60.

Hi aarushisingla,

Based on your approach, it does not look like you are properly accounting for 'duplicate' entries.

For example:
D - (first I) - G - (second I) - T

and

D - (second I) - G - (first I) - T

are the SAME result, so you are not supposed to count it twice (just once).

I'm going to approach the first part of your calculation in a different way - and then you can try to complete the overall calculation on your own:

- If the 2 I's are separated by 1 letter

There are only three possible placements for the two Is:

I _ I _ _
_ I _ I _
_ _ I _ I

You can arrange the remaining 3 letters (the D, G and T) in each option in 3! = 6 ways. So the total number of ways to have the two Is separated by just 1 space is 3(3!) = 18.

GMAT assassins aren't born, they're made,
Rich

Bunuel chetan2u generis VeritasKarishma

Why don't we divide 4! by 2! while we are calculating the number of cases with the 2 "I(s)" together just the way we did while calculating the total arrangements?

Thanks!

Hi,

There are total 5 letters, out of which 2 are same. So ways = 5!/2!=60, but this also includes when 2 Is are together.

So, we have to subtract the cases when two Is are together => For this take 2Is as 1 letter, That is D, (II), G and T.
And 4 letters can be arranged in 4! or 24 ways

Our answer = 60-24=36

ScottTargetTestPrep, why is it that the Is together is 4! = 24 instead of 4! * 2! = 48?? Don't the boxes of [4] [ii] [2] [1] need to be multiplied by 2! to account for the different orders of the Is?

Response:

If we were trying to find the number of arrangements of the letters D, I, G, I, and T, in which the letters G and T were together - in other words, if we were calculating the number of arrangements of [D] [I] [I] [G-T] - then you would be 100% correct that we would multiply the number of arrangements of these four items (two of which are indistinguishable) by 2! because [G-T] could appear as G-T or T-G. However, when we are arranging [I - I] [D] [G] [T], we should not multiply by 2!. There is no way to arrange I - I other than I - I because the two Is are identical.

Hey Scott, why is it not 4!/2! because two 'I's are also repeating?

Response:

It’s because here we are considering [I-I] as one single item instead of two items. Again, you can look at all 4 brackets ([I-I], [D], [G] and [T]) — they are all different. If two of these items were identical (for example, [I-I], [D], [G] and [I-I]), then we would indeed have to divide 4! by 2!; however, here, we don’t have two of these items that are identical.

Solution:

Total ways to arrange the letters of the word D,I,G,I,T =5!/2! =60

Number of ways the I's can be together =4! ways =24

Number of ways in which the I's are not together or in other words there is a letter b/w them

=60-24

=36 (option d)

Devmitra Sen
GMAT SME

Hi ScottTargetTestPrep:

How come we aren't doing this:

4! x 2?

Since I can be ordered

"I" really can't be ordered. Forget the GMAT, it's just 2 Is together, why would the order matter? For the same reason, we divide 60 by 2! when finding the total possible combinations

Hi everyone, i was wondering if you could explain a little bit more the example with the word google and if so how could we solve it the same way we solved the D,I,G,I,T example.

gmatophobia

If one "I" is at the start or end, the second "I" will have 3 places to select from, other letters will be placed as 3*2*1 thus 3*3*2*1 = 18 ways if I at start. Similarly, 18 ways if "I" is at the end. Thus total of 18+18 = 36 cases

If I is at any other place, then the second "I" can't be placed at 2 neighboring places thus it will have 2 places to select from and rest of the numbers will be placed as 3*2*1
Now this first I can be at second position, third position or fourth the above case II logic will remain same and thus 3*3*2*1=18

Thus total of 36 + 18 = 54 ways.
Can you please help me understand the mistake I made.

Rickooreo - You are double counting the cases -

Let's consider this scenario -



I _ _ _ _

The second "I" can occupy any of the three places except the highlighted one, and then you're arranging the remaining digits in 3! ways. However, note , in this logic, the second "I" can well occupy the last position. As a matter of fact, the 18 ways will include cases in which the second "I" is in the last place.

I _ _ _ I

Hence, this statement double counts.


IMO a better approach would be to consider all the cases and exclude those cases in which the two I are adjacent to each other.

Hope this helps.

When you calculate the number of ways the Is can be together (4!), why don't you divide by '2!'? Don't we want to remove the strings where I is repeating here too?

Hi skysailor
Thanks for your query.


To get a better understanding of the “why” behind the formula used, check out the example given below.


EXAMPLE:
How many 4-letter words can be formed using the letters of the word POOR?

One basic concept on permutations tells us that if we need to arrange 4 distinct elements, we can do so in 4! (= 24) ways. So, if all 4 letters were distinct, there would be 24 possible words that can be formed using those letters.

But, as you must have observed, in the word POOR, we have 2 identical letters. You must first understand that if we still do 4!, it will involve a lot of extra counting – that is it will count the same word more than once. See how:

1. Suppose that in the word POOR, we keep P and R fixed and just change the positions of the two Os. Then, will we get a new word? NO!
   
   • POOR is the same as POOR.
2. Similarly, let’s consider another arrangement of the letters, P, O, O, R, say, ROOP. Even now, ROOP is the same as ROOP.

Since ‘O’ is not different from ‘O’, it doesn’t make sense to count POOR and POOR as two different words, or ROOP and ROOP as two different words.

But when we calculate 4!, it does count these words as distinct. And hence, we need to factor these out.

Factor out repetitions:
Since for every fixed position of P and R, the two Os can be arranged in 2! ways, we will divide the total of 4! by 2!.

Note that had there been three Os, we would have divided it by 3! to factor out all arrangements obtained by ONLY arranging the Os. (Again, because these three Os can be arranged in 3! ways, for every fixed position of the other elements.)


To summarize the example, we can say the total number of words that can be formed using the letters of the word POOR = \(\frac{4! }{ 2!}\) = 12.


GENERAL PRINCIPLE
RESULT: If we must find the total number of arrangements of ‘n’ items, out of which

1. ‘p’ items are identical, then the total possible arrangements = n! / p!
2. ‘p’ items are identical to each other, and ‘q’ items are identical to each other, then the total possible arrangements = n!/(p!q!), and so on.

APPLICATION ON YOUR QUESTION:
Finally, let’s apply this to our question at hand.

We need to find the total number of possible 5-letter words that can be formed using the letters of the word DIGIT. So,

• Total letters available (n) = 5
• Repetitive letters count (p) = 2 (two Is)
  
  • Hence, the total number of words = 5! / 2! = 5 × 4 × 3 = 60.

Hope this helps!


Best,
Aditi Gupta
Quant expert, e-GMAT