chetan2u
In the 2nd case ie when the 2 Is are together why are we not dividing the 4! by 2!, as the interchange of the 2 Is when they are together will lead to double the number of arrangements...
Please correct me...
Thanks in advance

total number of ways that the letters can be arranged is (5!/2!) =60

it is easiest to find how many ways the I's can be together.

First we can treat both I's as a single entity ( I &I ).

This essentially this means we are arranging four entities which becomes 4!.

Thus the answer to the question is (5!/2!)- 4! = 36

We have letters: D, G, I, I, T
We need to have at least one letter between both the "I"s.
We can find the answer by finding: (Total Number of ways the 5 letters can be arranged) - (Number of ways the letters are arranged such that both the "I"s stay together) -> (a)

Number of ways the 5 letters can be arranged is \(\frac{5!}{2!}\) = 60 ways

To calculate the number of ways the letters can be arranged such that both the "I"s stay together, we need to consider both "I"s as one element (can't be separated). Thus, we only have 4 elements to arrange now- "D", "G", "II", "T".
These 4 elements can be arranged in 4! ways = 24 ways [We do not have to worry about interchanging the two "I"s positions are they are not distinct elements].

Thus, From (a), the required number of arrangements = 60 - 24 = 36 ways.

Answer D

Why isn’t 4 factorial divided by 2 factorial?

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Hi, Scott,

I do have a query regarding no. of ways to arrange I's together (I1 and I2). Why can not we write 4! * 2! as we can also arrange identical I's in two ways and that will give the option A (60-48 = 12).

Regards,
Raxit.

Hi A77777,

For the sake of calculating a permutation, the 'restriction' that the two "I"s must be 'next to one another' is essentially the same as saying that there is only one "I."

With 4 different letters, there are 4! = (4)(3)(2)(1) = 24 different arrangements

You can actually 'map' them all out if you choose; rather than list every option all at once, here are the 6 options that would start-off with the Is:

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

... and the 6 options that would start-off with the D:

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

There are two other 'sets of 6'; one that starts with a G and another that starts with a T.

GMAT assassins aren't born, they're made,
Rich
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Hi Raxit85,

You bring up a good question. Although the prompt does not explicitly state it, we are meant to assume that the two "I"s are identical. In simple terms, if you had just two letters (re: the two "I"s), then you could form just ONE distinct 'word': II (not two words - again, because the two "I"s are identical).

IF... the prompt did not want us to think in those terms, then we wouldn't have duplicate letters at all - it would just be 5 different letters and the question would ask something to the effect of "how many different ways are there to arrange the letters A, B, C, D and E so that there is at least one letter between the A and the B?"

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

they are indistinguishable, hence you do not multiply by 2.
So by your logic IIDGT and IIDGT are TWO different words, with the I's interchanged. No they are same. And therefore you club the I's together and take it as 1 entity.