My ans is d.
total no. arrangements= 7!/2
no. of arrangements in which two I are together is 6!

No. of arrangements in which I are sep. by atleast one character= 7!/2-6!= 1800

According to me

The total No of arrangements Possible (Even without NO character between 2 'I's) = 7! /(2! 2!) = 1260

Now the total arrangements where 2 'I's are TOGETHER = 6!/2! = 360

So the total arrangements where 2 'I' are NOT TOGETHER = 1260-360 = 900 Answer C

Correct me if I am wrong.

sandeep_chhabra wrote:According to me

The total No of arrangements Possible (Even without NO character between 2 'I's) = 7! /(2! 2!) = 1260

Now the total arrangements where 2 'I's are TOGETHER = 6!/2! = 360

So the total arrangements where 2 'I' are NOT TOGETHER = 1260-360 = 900 Answer C

Correct me if I am wrong.

I think you are correct. I guess the 2!2! is because you have 2T and 2I.

Could you please explain what it the 'theoretical' formula to be used?

For example: If I have 4 letters A/B/C/C --> total number of arrangements if 4!/2!= 12 ?

Thanks

sandeep_chhabra wrote:According to me

The total No of arrangements Possible (Even without NO character between 2 'I's) = 7! /(2! 2!) = 1260

Now the total arrangements where 2 'I's are TOGETHER = 6!/2! = 360

So the total arrangements where 2 'I' are NOT TOGETHER = 1260-360 = 900 Answer C

Correct me if I am wrong.

Thank you for your answer. Could you explain how could you get to 7!/(2!2!) (I think it is 2!2! because of the double I and double T but I'm not sure) and how could you get the 61/2! (I don't have any clue how you reached here)

molt_llest wrote:

sandeep_chhabra wrote:According to me

The total No of arrangements Possible (Even without NO character between 2 'I's) = 7! /(2! 2!) = 1260

Now the total arrangements where 2 'I's are TOGETHER = 6!/2! = 360

So the total arrangements where 2 'I' are NOT TOGETHER = 1260-360 = 900 Answer C

Correct me if I am wrong.

Thank you for your answer. Could you explain how could you get to 7!/(2!2!) (I think it is 2!2! because of the double I and double T but I'm not sure) and how could you get the 61/2! (I don't have any clue how you reached here)

molt_llest you are absolutely right there. 7!/(2! 2!) - here 2! is for 2 'I's and another 2! is to 2 'T's.

for 6! - actually we considered the 2 'I' to be ONE ENTITY because we want them together and thus counted them as one so the remaining letters that we are left with are 6. Now we divide 6! by 2! because of double 'T's.

Hope i was able to explain it

sandeep_chhabra wrote:

molt_llest wrote:

sandeep_chhabra wrote:According to me

The total No of arrangements Possible (Even without NO character between 2 'I's) = 7! /(2! 2!) = 1260

Now the total arrangements where 2 'I's are TOGETHER = 6!/2! = 360

So the total arrangements where 2 'I' are NOT TOGETHER = 1260-360 = 900 Answer C

Correct me if I am wrong.

Thank you for your answer. Could you explain how could you get to 7!/(2!2!) (I think it is 2!2! because of the double I and double T but I'm not sure) and how could you get the 61/2! (I don't have any clue how you reached here)

molt_llest you are absolutely right there. 7!/(2! 2!) - here 2! is for 2 'I's and another 2! is to 2 'T's.

for 6! - actually we considered the 2 'I' to be ONE ENTITY because we want them together and thus counted them as one so the remaining letters that we are left with are 6. Now we divide 6! by 2! because of double 'T's.

Hope i was able to explain it

Perfect!!! thank you very much, now is all clear.

The generic forumla for word jumble questions is:

n!/r!s!t!...

in which n is the total number of letters and r, s, t, etc... are the number of duplicate letters.

For example:

How many different "words" can be made out of:

DESERT = 6!/2!

DESSERT = 7!/2!2!

DESSERTS = 8!/3!2!

DESSERTED = 9!/3!2!2! (hey, it's real word! "Last night after the main course, I got desserted by the waiter." )

Of course, this formula applies to all permutation questions in which you use all the entities and not all of them are distinct.

For example:

In how many different ways can 3 pennies, 2 dimes and 2 quarters be arranged, if coins of the same type are identical?

Answer: 7!/3!2!2!