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Find the number of ways in which 4 letters may be selected

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Find the number of ways in which 4 letters may be selected  [#permalink]

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New post 11 Jan 2012, 18:55
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Question Stats:

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Find the number of ways in which 4 letters may be selected from the word "Examination"?

A. 66
B. 70
C. 136
D. 330
E. 4264

I was getting 142 but it is wrong,can anyone please help with this....??????

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Re: Combinations  [#permalink]

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New post 26 Jan 2012, 10:02
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1
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raanan wrote:
I would be thankful if someone could point out what is wrong in my solution:
In total we have to select 4 out of 11, so its 11C4.
each of the 3 couples can be substituted within itself, so for each we have 2!=2 inner combinations.
counting out those duplicities: 11C4/(2!*2!*2!) = 41.25 :-(


Find the number of ways in which 4 letters may be selected from the word "Examination"?
A. 66
B. 70
C. 136
D. 330
E. 4264

We have, 11 letters: {A, A, E, I, I, O, M, N, N, T, X};
Out of them there are 3 pairs: {A, A}, {I, I} and {N, N}.
So, # of distinct letters is 8: {A, E, I, I, O, M, N, N, T, X}

As pointed out, there are 3 different cases possible.
{abcd} - all 4 letters are distinct: \(C^4_8=70\);
{aabc} - two letters are alike and other two are distinct: \(C^1_3*C^2_7=63\) (\(C^1_3\) is a # of ways to choose which two letters will be alike from 3 pairs and \(C^2_7\) # of ways to choose other two distinct letters from 7 letters which are left);
{aabb} - two letters are alike and other two letters are also alike: \(C^2_3=3\) (\(C^3_3\) is a # of ways of choosing two pairs of alike letter from 3 such pairs);

Total=70+63+3=136.

Answer: C.

So, as you can see you can not just pick 4 letters with \(C^4_{11}\) and then divide it by some factorial as there are 3 different cases possible and each has its own factorial correction.

Hope it's clear.
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Re: Combinations  [#permalink]

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New post 11 Jan 2012, 21:32
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EXAMINATION has 11 letters, and in which 'A', 'I' and 'N', all occur twice. 11C4, would have been fine if all letters were distinct.

So, we have E, X, M, T, O, (AA), (II), (NN). 8 distinct letters.

1. 4 letters selected, which are all distinct: 8C4 = 70
2. 2 letters alike, and 2 distinct (eg: AAEX) = 3C1 x 7C2 = 63
3. 2 letters alike, and 2 letters alike (eg: AAII) = 3C2 = 3

So answer is, 70 + 63 + 3 = 136.

The tricky part is getting the second one correct, ( 2 letters alike, and 2 distinct)

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Re: Combinations  [#permalink]

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New post 11 Jan 2012, 20:13
Hi there, I got 330 and don't understand why this would be wrong.
"examination" has 11 letters out of which we are selecting 4 letters.
11C4=11!/4!7!=330. What's the source of this question?
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Re: Combinations  [#permalink]

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New post 26 Jan 2012, 07:41
I would be thankful if someone could point out what is wrong in my solution:
In total we have to select 4 out of 11, so its 11C4.
each of the 3 couples can be substituted within itself, so for each we have 2!=2 inner combinations.
counting out those duplicities: 11C4/(2!*2!*2!) = 41.25 :-(
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Re: Find the number of ways in which 4 letters may be selected  [#permalink]

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New post 30 Jun 2013, 23:59
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

*New project from GMAT Club!!! Check HERE

Theory on Combinations: math-combinatorics-87345.html

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PS questions on Combinations: search.php?search_id=tag&tag_id=52

Tough and tricky questions on Combinations: hardest-area-questions-probability-and-combinations-101361.html

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Re: Combinations  [#permalink]

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New post 24 Jul 2013, 13:01
Bunuel wrote:
raanan wrote:
I would be thankful if someone could point out what is wrong in my solution:
In total we have to select 4 out of 11, so its 11C4.
each of the 3 couples can be substituted within itself, so for each we have 2!=2 inner combinations.
counting out those duplicities: 11C4/(2!*2!*2!) = 41.25 :-(


Find the number of ways in which 4 letters may be selected from the word "Examination"?
A. 66
B. 70
C. 136
D. 330
E. 4264

We have, 11 letters: {A, A, E, I, I, O, M, N, N, T, X};
Out of them there are 3 pairs: {A, A}, {I, I} and {N, N}.
So, # of distinct letters is 8: {A, E, I, I, O, M, N, N, T, X}

As pointed out, there are 3 different cases possible.
{abcd} - all 4 letters are distinct: \(C^4_8=70\);
{aabc} - two letters are alike and other two are distinct: \(C^1_3*C^2_7=63\) (\(C^1_3\) is a # of ways to choose which two letters will be alike from 3 pairs and \(C^2_7\) # of ways to choose other two distinct letters from 7 letters which are left);
{aabb} - two letters are alike and other two letters are also alike: \(C^2_3=3\) (\(C^3_3\) is a # of ways of choosing two pairs of alike letter from 3 such pairs);

Total=70+63+3=136.

Answer: C.

So, as you can see you can not just pick 4 letters with \(C^4_{11}\) and then divide it by some factorial as there are 3 different cases possible and each has its own factorial correction.

Hope it's clear.



Hi Bunuel,
If the ques states that no repititions are possible, we will divide by factorial of no of time the letters are repeated to get the answer?Am i correct?Kindly clarify
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Re: Combinations  [#permalink]

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New post 25 Jul 2013, 01:29
up4gmat wrote:
Bunuel wrote:
raanan wrote:
I would be thankful if someone could point out what is wrong in my solution:
In total we have to select 4 out of 11, so its 11C4.
each of the 3 couples can be substituted within itself, so for each we have 2!=2 inner combinations.
counting out those duplicities: 11C4/(2!*2!*2!) = 41.25 :-(


Find the number of ways in which 4 letters may be selected from the word "Examination"?
A. 66
B. 70
C. 136
D. 330
E. 4264

We have, 11 letters: {A, A, E, I, I, O, M, N, N, T, X};
Out of them there are 3 pairs: {A, A}, {I, I} and {N, N}.
So, # of distinct letters is 8: {A, E, I, I, O, M, N, N, T, X}

As pointed out, there are 3 different cases possible.
{abcd} - all 4 letters are distinct: \(C^4_8=70\);
{aabc} - two letters are alike and other two are distinct: \(C^1_3*C^2_7=63\) (\(C^1_3\) is a # of ways to choose which two letters will be alike from 3 pairs and \(C^2_7\) # of ways to choose other two distinct letters from 7 letters which are left);
{aabb} - two letters are alike and other two letters are also alike: \(C^2_3=3\) (\(C^3_3\) is a # of ways of choosing two pairs of alike letter from 3 such pairs);

Total=70+63+3=136.

Answer: C.

So, as you can see you can not just pick 4 letters with \(C^4_{11}\) and then divide it by some factorial as there are 3 different cases possible and each has its own factorial correction.

Hope it's clear.



Hi Bunuel,
If the ques states that no repititions are possible, we will divide by factorial of no of time the letters are repeated to get the answer?Am i correct?Kindly clarify


In this case we would be basically selecting 4 letters out of 8: {A, E, I, O, M, N, T, X}.
Answer: \(C^4_8=70\)
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PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: Combinations  [#permalink]

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New post 30 May 2014, 04:50
Bunuel wrote:
raanan wrote:
I would be thankful if someone could point out what is wrong in my solution:
In total we have to select 4 out of 11, so its 11C4.
each of the 3 couples can be substituted within itself, so for each we have 2!=2 inner combinations.
counting out those duplicities: 11C4/(2!*2!*2!) = 41.25 :-(


Find the number of ways in which 4 letters may be selected from the word "Examination"?
A. 66
B. 70
C. 136
D. 330
E. 4264

We have, 11 letters: {A, A, E, I, I, O, M, N, N, T, X};
Out of them there are 3 pairs: {A, A}, {I, I} and {N, N}.
So, # of distinct letters is 8: {A, E, I, I, O, M, N, N, T, X}

As pointed out, there are 3 different cases possible.
{abcd} - all 4 letters are distinct: \(C^4_8=70\);
{aabc} - two letters are alike and other two are distinct: \(C^1_3*C^2_7=63\) (\(C^1_3\) is a # of ways to choose which two letters will be alike from 3 pairs and \(C^2_7\) # of ways to choose other two distinct letters from 7 letters which are left);
{aabb} - two letters are alike and other two letters are also alike: \(C^2_3=3\) (\(C^3_3\) is a # of ways of choosing two pairs of alike letter from 3 such pairs);

Total=70+63+3=136.

Answer: C.

So, as you can see you can not just pick 4 letters with \(C^4_{11}\) and then divide it by some factorial as there are 3 different cases possible and each has its own factorial correction.

Hope it's clear.


When is it ok to use the anagram method for such problems?

I did 11! / 5!2!2!2! , but obviously incorrect

Any idea?

Thanks!
Cheers
J
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Re: Combinations  [#permalink]

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New post 30 May 2014, 06:20
jlgdr wrote:
Bunuel wrote:
raanan wrote:
I would be thankful if someone could point out what is wrong in my solution:
In total we have to select 4 out of 11, so its 11C4.
each of the 3 couples can be substituted within itself, so for each we have 2!=2 inner combinations.
counting out those duplicities: 11C4/(2!*2!*2!) = 41.25 :-(


Find the number of ways in which 4 letters may be selected from the word "Examination"?
A. 66
B. 70
C. 136
D. 330
E. 4264

We have, 11 letters: {A, A, E, I, I, O, M, N, N, T, X};
Out of them there are 3 pairs: {A, A}, {I, I} and {N, N}.
So, # of distinct letters is 8: {A, E, I, I, O, M, N, N, T, X}

As pointed out, there are 3 different cases possible.
{abcd} - all 4 letters are distinct: \(C^4_8=70\);
{aabc} - two letters are alike and other two are distinct: \(C^1_3*C^2_7=63\) (\(C^1_3\) is a # of ways to choose which two letters will be alike from 3 pairs and \(C^2_7\) # of ways to choose other two distinct letters from 7 letters which are left);
{aabb} - two letters are alike and other two letters are also alike: \(C^2_3=3\) (\(C^3_3\) is a # of ways of choosing two pairs of alike letter from 3 such pairs);

Total=70+63+3=136.

Answer: C.

So, as you can see you can not just pick 4 letters with \(C^4_{11}\) and then divide it by some factorial as there are 3 different cases possible and each has its own factorial correction.

Hope it's clear.


When is it ok to use the anagram method for such problems?

I did 11! / 5!2!2!2! , but obviously incorrect

Any idea?

Thanks!
Cheers
J


11!/(2!2!2!) is the number of arrangements of the word "examination", which is not what the question is asking.
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: Find the number of ways in which 4 letters may be selected  [#permalink]

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New post 03 Jul 2015, 22:29
1
I think the problem people are having with this question is in deciphering: This is a combination question where order of 'picking' does not matter. Then why do I've to group these similar letters.

Well I had this problem too, this is how I solved it:
1. Understand what the question is asking: Is it asking about me 'picking' the letters? Or is it asking me about 'arranging' these letters differently?
Ans: Its asking about 'picking'. So there you go, 1 question down :)
2. DONOT forget that the formula \(nCr\) is for n 'different' things :) - So here, you HAVE TO group the similar letter into sets.
3. Ok, now you've grouped them, what should be done now?
Well Since, you've grouped them, you can have different cases of how u now 'pick' !
-> Now, refer to what Bunuel said in his solution, and it'll make sense :)

Hope this helps!!
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Re: Find the number of ways in which 4 letters may be selected  [#permalink]

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Re: Find the number of ways in which 4 letters may be selected &nbs [#permalink] 11 Aug 2018, 12:38
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