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

Question

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....??????

Bunuel · Accepted Answer

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.

kinkmetal · Answer

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)

Cheers...

Sistersunshine · Answer

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?

FuzzyBuzzard · Answer

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 :-(

Bunuel · Answer

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

DS questions on Combinations: search.php?search_id=tag&tag_id=31
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

up4gmat · Answer

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

Bunuel · Answer

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

jlgdr · Answer

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

Bunuel · Answer

11!/(2!2!2!) is the number of arrangements of the word "examination", which is not what the question is asking.

ddg · Answer

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!!

MathRevolution · Answer

@ EXAMINATION: Total letters: 11 -> A {2}, I{2}, N {2},

Letters to be selected are: 4

Possible cases:

Case I:   All 4 letters are different:  ^8{C_4}  = 70

CaseII:   Two letters same and two different:  ^3{C_1}  *  ^7{C_2}  = 3 * 21 = 63

CaseIII:   Two pairs of same letters:  ^3{C_2}  = 3

Total ways:  70  +  63  +  3  = 136

Answer C

samarpan.g28 · Answer

Hi there, hope this solution helps.

Kinshook · Answer

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

Letters available: - 
E-1
x-1
a-2
m-1
i-2
n-2
t-1
o-1

1 letters = {E,x,m,t,o} - 5 choices
2 letters = {a,i,n} - 3 choices

Case 1: All 4 letters are different
Number of ways to select 4 letters = 8C4 = 70 ways

Case 2: 2 letters are alike and 2 are different
Number of ways to select 4 letters = 3C1*7C2 = 3*21 = 63 ways

Case 3: 2 letters are alike for one kind and 2 letter are alike of another kind
Number of ways to select 4 letters = 3C2 = 3 ways

Number of ways to select 4 letters = 70 + 63 + 3 = 136

IMO C

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

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

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

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards