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by parveen110 » Sun Mar 02, 2014 4:34 am
In how many of the distinct permutations of the letters in the word MISSISSIPPI do the 4 I's not come together?

What is wrong in my approach?

# of ways s,s,s,s,p,p,m can be arranged: 7!/4!*2!
now, there are 8 places to insert four I's, that can be done in 8C4 ways.

To sum it up, 8C4*7!/4!*2!

Thank You.
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by Brent@GMATPrepNow » Sun Mar 02, 2014 6:30 am
parveen110 wrote:In how many of the distinct permutations of the letters in the word MISSISSIPPI do the 4 I's not come together?

When we want to arrange a group of items in which some of the items are identical, we can use something called the MISSISSIPPI rule. It goes like this:
If there are n objects where A of them are alike, another B of them are alike, another C of them are alike, and so on, then the total number of possible arrangements = n!/[(A!)(B!)(C!)....]

Okay, to solve the question, we must recognize the following:
# of "words" where the 4 I's are NOT together = (total # of arrangements) - (# of arrangements where the 4 I's ARE together)

total # of arrangements
Here, we're dealing with the letters M, I, I, I, I, S, S, S, S, P, and P
There are 11 letters in total
There are 4 identical I's
There are 4 identical S's
There are 2 identical P's
So, the total number of possible arrangements = 11!/[(4!)(4!)(2!)]
= 69,300

# of arrangements where the 4 I's ARE together
To ensure that all 4 I's are together, let's GLUE them together to form just ONE GIANT "LETTER" (IIII). Our letters are now as follows: M, S, S, S, S, P, P, and IIII
There are 8 letters in total
There are 4 identical S's
There are 2 identical P's
So, the total number of possible arrangements = 8!/[(4!)(2!)]
= 840

So, # of "words" where the 4 I's are NOT together = 69,300 - 840 = 68,460

Cheers,
Brent
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by Brent@GMATPrepNow » Sun Mar 02, 2014 6:33 am
parveen110 wrote: What is wrong in my approach?

# of ways s,s,s,s,p,p,m can be arranged: 7!/4!*2!
now, there are 8 places to insert four I's, that can be done in 8C4 ways.

To sum it up, 8C4*7!/4!*2!
I'm not sure what the part in green means.
What exactly is happening when you have 8 choose 4 (8C4)?
What are the 8 objects and what are you doing with the 4 that you choose?

Cheers,
Brent
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by parveen110 » Sun Mar 02, 2014 7:43 pm
Brent@GMATPrepNow wrote:
parveen110 wrote: What is wrong in my approach?

# of ways s,s,s,s,p,p,m can be arranged: 7!/4!*2!
now, there are 8 places to insert four I's, that can be done in 8C4 ways.

To sum it up, 8C4*7!/4!*2!
I'm not sure what the part in green means.
What exactly is happening when you have 8 choose 4 (8C4)?
What are the 8 objects and what are you doing with the 4 that you choose?

Cheers,
Brent
Hi Brent,

Thank you for the reply. The text in the green means the following:

I may arrange all the letters except 4 I's in 7!/4!*2!.
Let's say, one of the arrangements may be,
_s_s_s_s_p_p_m_

Now that leaves 8 spaces to arrange 4 I's. Therefore, I may choose any of the four places out of eight in 8C4 ways.

So, total # of arrangements of the word so that all the 4 I's are never together, 8C4*7!/4!*2!.

I hope I have made myself clear. Please clarify the flaw in my approach.

Thanks!
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by GMATGuruNY » Mon Mar 03, 2014 6:41 am
parveen110 wrote:
Brent@GMATPrepNow wrote:
parveen110 wrote: What is wrong in my approach?

# of ways s,s,s,s,p,p,m can be arranged: 7!/4!*2!
now, there are 8 places to insert four I's, that can be done in 8C4 ways.

To sum it up, 8C4*7!/4!*2!
I'm not sure what the part in green means.
What exactly is happening when you have 8 choose 4 (8C4)?
What are the 8 objects and what are you doing with the 4 that you choose?

Cheers,
Brent
Hi Brent,

Thank you for the reply. The text in the green means the following:

I may arrange all the letters except 4 I's in 7!/4!*2!.
Let's say, one of the arrangements may be,
_s_s_s_s_p_p_m_

Now that leaves 8 spaces to arrange 4 I's. Therefore, I may choose any of the four places out of eight in 8C4 ways.

So, total # of arrangements of the word so that all the 4 I's are never together, 8C4*7!/4!*2!.

I hope I have made myself clear. Please clarify the flaw in my approach.

Thanks!
The intent of the problem is not crystal clear.
Brent has offered a solution for the following:
How many ways can the letters in MISSISSIPPI be arranged if the four I's cannot occupy four adjacent positions?
Here, it's ok for TWO OR THREE I's to occupy adjacent positions, but it's NOT ok for ALL FOUR I's to occupy adjacent positions.
You have offered a solution for the following:
How many ways can the letters in MISSISSIPPI be arranged if no two I'S can occupy adjacent positions?
Here, it's NOT ok for two or three I's to occupy adjacent positions; there must be at least one consonant between any two successive I's.
Your solution for the latter problem is correct.
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