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Slot Method Only Please

by knight247 » Sun Oct 02, 2011 9:48 am
How many words can be formed by taking 4 letters at a time out of the letters of the word MATHEMATICS?

OA is 2454. I'm kinda familiar with solving using combinatorics. Hoping to get a solution using ONLY the slot method. Detailed explanations would be appreciated
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by shankar.ashwin » Sun Oct 02, 2011 11:00 am
I am not sure if this is the slot method, but just the way I did it;

4 letter word can be formed in 3 diff ways
1) 4 diff letters
2) 3 diff letters
3) 2 diff letters

MATHEMATICS has 11 letters with 2M's, 2A's and 2T's

Case 1 _ _ _ _ (neglecting duplicates 8 choices)

8*7*6*5 = 1680

Case 2 ( 3 diff letters - 1 from letters that repeat and 2 from remaining)

3C1 * 7C2 * 4!/2! = 756

Case 3 (2 letter from letters that repeat)

3C2 * 4!/2!*2! = 18

Total would be 2454
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by knight247 » Sun Oct 02, 2011 11:05 am
Thats not slot method Ashwin. But anyways, thanks for taking the time.
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by shankar.ashwin » Sun Oct 02, 2011 11:10 am
I thought so too, so used to a fixed way of doing these problems, just not able to think any other way I guess :)
knight247 wrote:Thats not slot method Ashwin. But anyways, thanks for taking the time.
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by GmatMathPro » Sun Oct 02, 2011 11:49 am
I'm afraid you may be out of luck if you're hoping for a simple, straightforward slot method application on this one. If you were going to do the slot method, you would want four slots for the four letters: _ _ _ _. You could start with 11*10*9*8, but you HAVE to do something about the repetitions. Normally, if you were just permuting the whole word you would divide by 2!2!2! because there are three cases of two letters being repeat. The problem here is that some of the cases will have repetitions and some won't, like HECS is counted once, but MCMS is counted twice, and MMHH will be counted four times. Counting problems with repetitions work out very nicely when the repetitions are distributed evenly among the unique sequences. For example, how many teams of three can be chosen from 4 people, ABCD. If we do slot method here: 4*3*2=24, each unique group is counted 6 times, because each group of three can be sequenced in 3*2*1 ways. The repetitions are distributed evenly in that each unique string is represented exactly 6 times, so its easy to get the right answer by dividing by 6. With your problem, some of them will repeat, and some won't.

Let's take a look at a simpler example: How many two-letter sequences can you make from the word LOOK. Ignoring repeats for now, there should be 4*3=12 sequences. Let's list them out:

LO OL OL KL

LO OO OO KO

LK OK OK KO

LO, OO, OK, KO, OL all appear twice, and LK, KL only appear once. So there are 7 unique sequences. In the slot method you just multiply numbers together, but what could you possibly multiply together to get 7 that would make sense? Also, there's no way to divide 12 by any integer to get 7.

In your problem, 11*10*9*8=7920. If there were some repeat factor that we could divide out,x, then 7920/x=2454 should have an integer solution, but it doesn't: x=3.2273....

So, I don't think a traditional, straightforward slot method is going to work here.

So, how do you do it at all? I thought shankar.ashwin's method was good. On the more complicated counting problems sometimes you're just stuck breaking it into cases. The problem with the slot method is that it's "blind". It works best when everything from the pool you're picking from is created equal. But in this problem the M, A, and T wreak havoc, and there's no obvious way to correct it.
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by knight247 » Sun Oct 02, 2011 12:30 pm
Hey Pete,
Appreciate ur response bro. Ok how about I try it with the slot method and then u can tell me where I'm going wrong.

(A)MATHEMATICS= 11 letters

(B)MATHEICS=8 Letters i.e # of letters excluding the repeated elements

(C)MMAATT=6 Letters=3 pairs i.e. # of pairs


Lets consider the different scenarios
(1) All different i.e. ABCD. For this we look at (B)
For the 1st letter we have 8 choices
For the 2nd letter we have 7
For the 3rd letter we have 6
For the 4th letter we have 5
Total=8*7*6*5=1680

(2)Two same and two different i.e. AABC
We consider the first two slots as one. Now we look at B and can fill these two slots in 3 ways.
We consider the 3rd slot. We can fill this in 7 ways.
In the 4th slot, we have 6 ways
Total=3*7*6=126 (I know there is another multiplication with 6 that i'm missing here as when I do it with combinatorics I get 756)

(3)Two same and Two same i.e. AABB
We consider the first two slots as one. Now we can fill this slot in 3 ways...(Look at B)
We consider the next two slots as one. We can fill this slot in 2 ways...
Total=2*3=6 .........(I'm missing a multiplication by 3 here coz I get 18 when doing this by combinatorics)

Sum=1680+126+6=1812

I know u've already made a very comprehensive post. But would appreciate if u could comment on any possible way of improvising on the above method.
Last edited by knight247 on Sun Oct 02, 2011 1:10 pm, edited 1 time in total.
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by GmatMathPro » Sun Oct 02, 2011 12:37 pm
I think your problem is that when you "consider two slots as one", you force the two repeats to be together, so your 6 cases would be MMTT, MMAA, TTMM, TTAA, AATT, AAMM. But you're not counting ones like AMAM.
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by knight247 » Sun Oct 02, 2011 12:44 pm
I see. So I multiply scenario 3 with 4!/(2!2!) which equals 6. But then I get 6*6=36 when I should be getting 18.

Also, what about scenario 2?
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by GmatMathPro » Sun Oct 02, 2011 1:00 pm
Right. That's because in your original group of 6 everything is counted twice. What I mean is, when you multiply by 4!/(2!2!), you're saying, each of these 6 groups can each be arranged 6 ways. But if you apply that to both AAMM and MMAA, you're getting duplicates. Hence, it's two times too big. I'm pretty sure Scenario 2 is suffering from a similar problem. Play with it a little while and I'll take another look.
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by GmatMathPro » Sun Oct 02, 2011 1:24 pm
With scenario 2 you also have the problem that you're forcing the double letter to appear in the first slot: AABC, whereas it could be BAAC or BCAA, plus the problem of omitting things like ABCA.

So you could correct that by multiplying by 4!/2!=12, but now we're double counting other sequences too. For example, 1 of the 126 sequences that your method produces is MMEC, another would be MMCE. So we get the same problem as scenario 3 where we're counting the rearrangements of the same group of letters twice. So you could correct this by dividing by 2.
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by GMATGuruNY » Sun Oct 02, 2011 5:30 pm
knight247 wrote:How many words can be formed by taking 4 letters at a time out of the letters of the word MATHEMATICS?

OA is 2454. I'm kinda familiar with solving using combinatorics. Hoping to get a solution using ONLY the slot method. Detailed explanations would be appreciated
4 different letters:
Number of options for the first letter = 8.
Number of remaining options for the second letter = 7.
Number of remaining options for the third letter = 6.
Number of remaining options for the fourth letter = 5.
To combine these options, we multiply:
8*7*6*5 = 1680.

3 different letters:
Number options for the pair of duplicate letters = 3. (MM, AA, or TT)
This pair must occupy a combination of 2 positions in the 4-letter word.
Number combinations of 2 that can be formed from 4 options = (4*3)/(2*1) = 6.
Number of remaining options for the next letter = 7. (Any of the 8 letters but the one already used.)
Number of remaining options for the last letter = 6. (Any of the 8 letters but the 2 already used.)
To combine these options, we multiply:
3*6*7*6 = 756.

2 different letters:
The 4 positions in the word can be divided into pairs as follows:
1,2....3,4
1,3....2,4
1,4....2,3
Thus, the number of ways to divide the 4-letter word into 2 pairs = 3.
Number of options for the first pair of duplicate letters = 3. (MM, AA, or TT)
Number of remaining options for the second pair of duplicate letters = 2. (Of MM, AA, and TT, any but the pair already used).
To combine these options, we multiply:
3*3*2*1 = 18.

Total options = 1680+756+18 = 2454.
Last edited by GMATGuruNY on Mon Oct 03, 2011 1:48 am, edited 1 time in total.
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by knight247 » Sun Oct 02, 2011 10:51 pm
GMATGuruNY wrote:2 different letters:
Number of options for the first pair of duplicate letters = 3. (MM, AA, or TT)
Number of remaining options for the second pair of duplicate letters = 2. (Of MM, AA, and TT, any but the pair already used).
The first pair of duplicate letters must occupy a combination of 2 positions in the 4-letter word.
Number of combinations of 2 that be formed from 4 options = (4*3)/(2*1) = 6.
The second pair of duplicate letters must occupy the remaining 2 positions in the 4-letter word.
Number of combinations of 2 that be formed from the remaining 2 positions = (2*1)/(2*1) = 1.
To combine these options, we multiply:
3*2*6*1 = 18.
Hey Mitch,
I think there is an error in ur last part. 3*2*6*1=36 but u've mentioned it as 18. Looks like that part still needs some work.lol.
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by GMATGuruNY » Mon Oct 03, 2011 2:25 am
Thanks. I've amended the post.

In the amended post, I've written out the ways in which the 4 positions in the word can be divided into pairs, since there are only 3 options:
1,2...3,4
1,3...2,4
1,4...2,3

Here's how the slot method could be used to count the number of ways in which the 4 positions could be DIVIDED into pairs:
The first position can be paired with any of the 3 other positions.
Thus, the number of options for the first position = 3.
2 positions left.
These remaining 2 positions must be paired with each other.
Thus, the number of options for the last 2 positions = 1.
To combine these options, we multiply:
3*1 = 3.

Other problems about DIVIDING elements into groups:

https://www.beatthegmat.com/forming-teams-t73034.html

https://www.beatthegmat.com/combination-t92104.html
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by seema19 » Mon Oct 03, 2011 6:50 am
shankar.ashwin wrote:I am not sure if this is the slot method, but just the way I did it;

4 letter word can be formed in 3 diff ways
1) 4 diff letters
2) 3 diff letters
3) 2 diff letters

MATHEMATICS has 11 letters with 2M's, 2A's and 2T's

Case 1 _ _ _ _ (neglecting duplicates 8 choices)

8*7*6*5 = 1680

Case 2 ( 3 diff letters - 1 from letters that repeat and 2 from remaining)

3C1 * 7C2 * 4!/2! = 756

Case 3 (2 letter from letters that repeat)

3C2 * 4!/2!*2! = 18

Total would be 2454
hi shankar.ashwin, can you plz explain the last 2 cases..
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by shankar.ashwin » Mon Oct 03, 2011 7:05 am
@seema19

Case 2 ( 3 diff letters - 1 from letters that repeat and 2 from remaining)

Here I am going to use only 3 letters to form a 4 letter word, one of the letters will be repeated. (for eg, consider arrangements of MMEI)

There are 3 choices to pick of the letter which is repeated from M,A and T

So thats 3C1.

After you pick 1, you have 7 distinct letters left (leave out the duplicates). From 7 i pick 2.

So thats 7C2.

4!/2! is all arrangements of these letters (dividing it by 2! bcos 1 letter is repeated)

So you have,

3C1 * 7C2 * 4!/2! = 756

Similarly for Case 3,

I pick only from letters which are repeated (only 3 choices, I pick 2)

3C2 and 4!/2!*2! is arrangements with 2 duplicates.

These are the only 3 cases which are possible to from a 4 letter word.
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