THEORY:

Permutations of \(n\) things of which \(P_1\) are alike of one kind, \(P_2\) are alike of second kind, \(P_3\) are alike of third kind ... \(P_r\) are alike of \(r_{th}\) kind such that: \(P_1+P_2+P_3+..+P_r=n\) is:

\(\frac{n!}{P_1!*P_2!*P_3!*...*P_r!}\).

For example number of permutation of the letters of the word "gmatclub" is 8! as there are 8 DISTINCT letters in this word.

Number of permutation of the letters of the word "google" is \(\frac{6!}{2!2!}\), as there are 6 letters out of which "g" and "o" are represented twice.

Number of permutation of 9 balls out of which 4 are red, 3 green and 2 blue, would be \(\frac{9!}{4!3!2!}\).

For more check the links below:

Combinatorics Made Easy!

Theory on Combinations

DS questions on Combinations
PS questions on Combinations

Tough and tricky questions on Combinations

So, as explained above, the number of arrangements of 5 letters D, G, I, I , and T out of which there are two identical I's will be 5!/2! = 60.
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Another approach.

Take the task of arranging the 5 letters and break it into stages.

Stage 1: Arrange the 3 CONSONANTS (D, G and T) in a row
We can arrange n unique objects in n! ways.
So, we can arrange the 3 consonants in 3! ways (= 6 ways)
So, we can complete stage 1 in 6 ways

IMPORTANT: For each arrangement of 3 consonants, there are 4 places where the two I's can be placed.
For example, in the arrangement DTG, we can add spaces as follows _D_T_G_
So, if we place each I in one of the available spaces, we can ENSURE that the two I's are never together.

Stage 2: Select two available spaces and place an I in each space.
Since the order in which we select the two spaces does not matter, we can use combinations.
We can select 2 spaces from 4 spaces in 4C2 ways (= 6 ways)
So we can complete stage 2 in 6 ways.

By the Fundamental Counting Principle (FCP), we can complete the 2 stages (and thus arrange all 5 letters) in (6)(6) ways (= 36 ways)

Answer: D

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. So be sure to learn this technique.

RELATED VIDEOS





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IMO 36

Total no of ways arranging 5 letter with one letter redundant is 5!/2!=60
No of ways two I's can be together 4!=24

no of ways at least one alpha is between two I's =60-24=36

Let us calculate the total ways. Those would be (5!/2!) = 60.

Now since the question says "at least" let us find the number of arrangements when both I's are together. (Tie them up). so we have 4! ways to arrange such that I's always come together. 4! = 24

60 - 24 = 36.

D is the answer.

Total ways to arrange these letters=5!/2!

Ways to arrange the letters if both I are together= 4!

Ways to arrange with both Is not together= 5!/2!-4!= 36

D is the answer
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Required: atleast one letter between two Is

Total cases = 5!/2! = 60
Total cases in which Is will be together = 4! = 24 {Treat both Is as one}

Hence total cases in which there is atleast one letter between the Is = 60 - 24 = 36

Correct Option: D

The requirement is 5 letter strings in which two I's are not together, they are separated by atleast one letter.

Consider this as below,

(Total number of 5 letter strings that can be formed) - (Total number of 5 letter strings in which two I's are together)
=(5!/2!) - (4!)
= 5*4*3 - 4*3*2
= 60 - 24
= 36

Finding total number of 5 letter strings - DIGIT = 5!/2! (2! because of the repetition of I's)
Finding total number of 5 letter strings with I's together - II_ _ _ = 4!

A couple of video explanations for this question:

"Direct" Method:



"Indirect" Method:

Could someone help me understand how the total number of cases is 5!/2!? I don't understand the reasoning behind that.

1. It is an ordering, therefore a permutation problem
2. There is a constraint
3. The constraint is an "Atleast One" constraint. so the opposite of it "none" is easier to find
4. Total number of permutations is 5!/2!=60 ways since I repeats twice
5. Number of permutations where I's are together is 3!*4 = 24
6. Permutations with constraint is (4)-(5)=36
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The approach i used was...I combined both I's together,so now there are four letters which can be arranged in 4! ways.

Total number of ways 5 letters can be arranged without any constraints is 5!.

So the the answer should be 5!-4! as ...can someone help to explain what wrong i did?

Hi

You are wrong in coloured portion..
When there are TWO similar letters, number of ways is 5!/2!=60..
and is 60-24=36
D
Hope it helps

This is a permutation problem because the order of the letters matters. Let’s first determine in how many ways we can arrange the letters. Since there are 2 repeating Is, we can arrange the letters in 5!/2! = 120/2 = 60 ways.

We also have the following equation:

60 = (number of ways to arrange the letters with the Is together) + (number of ways without the Is together).

Let’s determine the number of ways to arrange the letters with the Is together.

We have: [I-I] [D] [G] [T]

We see that with the Is together, we have 4! = 24 ways to arrange the letters.

Thus, the number of ways to arrange the letters without the Is together (i.e., with the Is separated) is 60 - 24 = 36.

Answer: D
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Hi All,

Since the two "I"s cannot be side-by-side, there are a limited number of ways to arrange the 5 letters. As such, with a little permutation math and some 'brute force', we can map out the possibilities:

_ _ _ _ _

If the first letter is an I, then the second letter CANNOT be an I (it would have to be one of the other 3 non-I letters)...

i 3

From here, any of the remaining letters can be in the 3rd spot. After placing one, either of the remaining two letters can be in the 4th spot and the last letter would be in the 5th spot...

i 3 3 2 1

This would give us (3)(3)(2)(1) = 18 possible arrangements with an I in the 1st spot.

If a non-I is in the 1st spot and an I is in the 2nd spot, then we have...

3 i _ _ _

A non-I would have to be in the 3rd spot, then either remaining letter could be 4th...

3 i 2 2 1

This would give us (3)(2)(2)(1) = 12 possible arrangements

Next, we could have two non-Is to start off, then Is in the 3rd and 5th spots...

3 2 i 1 i

This would give us (3)(2)(1) = 6 possible arrangements

There are no other options to account for, so we have 18+12+6 total arrangements.

Final Answer:

GMAT assassins aren't born, they're made,
Rich
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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 addition to that D, G, and T, could be arranged in 3! ways, so total = 4*3! = 4!.

Or, consider two I's as one unit {II}. We'll have 4 units: {D}, {G}, {T}, and {II}. The number of arrangements is 4!.

Hope it's clear.
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When answering counting question, one should always consider whether a useful approach would be listing and counting the possible outcomes.

How do we know when this approach may be the best approach?
If you check the answer choices (ALWAYS check the answer choices BEFORE performing any calculations) and find that the answer choices are relatively small (as they are here), then listing and counting might be a great approach.
Alternatively, if you have no idea how to solve the question using traditional counting techniques, you might again consider listing and counting.
Also, even if the answer choices are relatively large, you might start listing some outcomes, and while doing so, you might see a pattern emerge that you can use to determine the correct answer.

TLDR: Don't dismiss listing and counting as a possible approach to answering counting questions.

POSSIBLE OUTCOMES
Words in the form _ I _ I _
DIGIT
DITIG
GITID
GIDIT
TIGID
TIDIG
Words in the form _ I _ _ I
DIGTI
DITGI
GIDTI
GITDI
TIDGI
TIGDI
Words in the form _ _ I _ I
DGITI
DTIGI
GDITI
GTIDI
TDIGI
TGIDI
Words in the form I _ I_ _
ITIGD
ITIDG
IDITG
IDIGT
IGITD
IGIDT
Words in the form I _ _ I_
IDGIT
IGDIT
ITGID
IGTID
IDTIG
ITDIG
Words in the form I _ _ _ I
IDGTI
IDTGI
IGDTI
IGTDI
ITDGI
ITGDI

Done!
Answer = 36 = D

Did that take under 2 minutes? Probably.

Cheers,
Brent
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The correct answer is Choice D.

First, determine the total number of ways of rearranging the letters in DIGIT.

5! / 2! (accounts for the repetition of the 2 letter "I"s) = (5 x 4 x 3 x 2) / 2 = 120 / 2 = 60

Next, consider all the ways two letter "I"s can be adjacent within the word. They can either be in the 1/2 slot, the 2/3 slot, the 3/4 slot, or the 4/5 slot (4 locations), and for each of those 4 locations there are 3 x 2 x 1 = 6 other ways of rearranging the final 3 letters, so multiply 6 by 4 to get 24.

Subtract those 24 instances from the total to get 60 - 24 = 36

-Brian
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can you please explain why the No of ways two I's can be together is 4!=24?
thank you!!