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Permutation - Two alike and other two distinct problem

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Permutation - Two alike and other two distinct problem

by AndyB » Tue Mar 15, 2011 2:33 am
Hi All,

Need help in solving the below problem.
Q)Find the number of ways in which a selection of 4 letters can be made from the letters of the word "DISTILLATIONS".
a)2 alike, the other two distinct.

Ans)There are 13 letters of 8 different sorts I,I,,I,S,S,T,T,L,L,A,O,N,D.

My book gives the solution as -
The alike pair can be selected in 4 wasys.(Agreed)
While the other 2 distinct letters can be selected form the 7 distinct letters in 7C2 ways.
Hence the req number of ways are 4 x 7C2.

But i feel the solution is missing the point when (I,I) 2 alike is selected; we can still select the other 2 distinct from
(I,S,T,,L,A,O,N,D) eitht letters.So I think the solution given is wrong.

Could anyone please correct me if my understanding of the peoblem is wrong or explain me the correct answer.

Regards,

AndyB.
Last edited by AndyB on Tue Mar 15, 2011 6:49 am, edited 1 time in total.
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by HSPA » Tue Mar 15, 2011 2:58 am
I,I,,I,S,S,T,T,L,L,A,O,N,D
Q) 2 alike, the other two distinct.

two similar elements from I and there are 8 distinct elements left(I,S,T,L,A,O,N,D) : 3c2 * 8c2
two similar S elements and the rest distinct elements : 2c2*7c2
similar for T: 2c2* 7c2
similarly for L...

3(2c2*7c2) + 3c2*8c2 = 147 or 3(2c2*7c2)+ 3c2*7c2 = 126

Kindly post the OA
Last edited by HSPA on Tue Mar 15, 2011 4:53 am, edited 1 time in total.
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by srcc25anu » Tue Mar 15, 2011 4:49 am
DISTILLATION is 12 letter word (if there's no spelling error in posting the Q)
III TT LL D S A O N

2 identical words can be selected in 3 ways (either II, TT or DD)
remaining 2 letters can be selected out of 7 alphabets in 7c2 ways

hence 4 letter word with 2 same-2 distinct letters would be 3*7c2

If going by the Q and taking it as a 13 letter word with III TT LL SS D A O N
then identical words can be selected in 4 ways (either II, TT, SS or DD)
remaining 2 letters can be selected out of 7 alphabets in 7c2 ways

hence 4 letter word with 2 same-2 distinct letters would be 4*7c2
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by AndyB » Tue Mar 15, 2011 6:51 am
The word actually is "DISTILLATIONS"
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by GMATGuruNY » Tue Mar 15, 2011 7:07 am
AndyB wrote:Hi All,

Need help in solving the below problem.
Q)Find the number of ways in which a selection of 4 letters can be made from the letters of the word "DISTILLATION".
a)2 alike, the other two distinct.

Ans)There are 13 letters of 8 different sorts I,I,,I,S,S,T,T,L,L,A,O,N,D.

My book gives the solution as -
The alike pair can be selected in 4 wasys.(Agreed)
While the other 2 distinct letters can be selected form the 7 distinct letters in 7C2 ways.
Hence the req number of ways are 4 x 7C2.

But i feel the solution is missing the point when (I,I) 2 alike is selected; we can still select the other 2 distinct from
(I,S,T,,L,A,O,N,D) eitht letters.So I think the solution given is wrong.

Could anyone please correct me if my understanding of the peoblem is wrong or explain me the correct answer.

Regards,

AndyB.
2 identical letters:
III, TT, LL, SS give us 4 choices.

2 distinct letters:
From III, TT, LL, SS: Since 1 of these 4 letters was used above, we have 3 choices left.
The remaining 4 letters (D, A, O, N) give us another 4 choices.
Total number of choices = 3+4 = 7.
Number of combinations of 2 that can be formed from 7 choices = 7C2 = 21.

To combine our choices for the identical pair with our choices for the distinct pair, we multiply the results above:
4*21 = 84.
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by AndyB » Tue Mar 15, 2011 7:19 am
Hi HSPA,

Sorry, I am not sure about the OA.

I think the ans could be:

When u consider the three alike pairs SS,TT,LL and other 7 letters, these can be done in:

3C1 x 7C2 ways.

When you consider of selecting II as the like pair there is a small change:

4C1 x 8C2

Total = 3C1 x 7C2 + 4C1 x 8C2 = 154.

But nost sure whether this ans is correct.
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by AndyB » Tue Mar 15, 2011 7:25 am
Hi Mitch,

The OA is the same given by you.

But my doubt is when we select I,I as the identical pair.
One extra I is left and then we can select 2 distinct letters from 8 letters.

At this point I get confused.Could you please help me in clarifying this confusion when I is selected as the like pair.

Regards,
AndyB.
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by GMATGuruNY » Tue Mar 15, 2011 7:28 am
AndyB wrote:Hi GmatGuru,

The OA is the same given by you.

But my doubt is when we select I,I as the identical pair.
One extra I is left and then we can select 2 distinct letters from 8 letters.

At this point I get confused.Could you please help me in clarifying this confusion when I is selected as the like pair.

Regards,
AndyB.
If II is chosen, the remaining I cannot be used because then the result would include 3 I's.
Thus, the distinct pair must be formed from the remaining 7 letters: S, T, L, D, A, O, N.
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by force5 » Tue Mar 15, 2011 7:33 am
2 identical letters:
III, TT, LL, SS give us 4 choices.

2 distinct letters:
From III, TT, LL, SS: Since 1 of these 4 letters was used above, we have 3 choices left.
The remaining 4 letters (D, A, O, N) give us another 4 choices.
Total number of choices = 3+4 = 7.
Number of combinations of 2 that can be formed from 7 choices = 7C2 = 21.

To combine our choices for the identical pair with our choices for the distinct pair, we multiply the results above:
4*21 = 84.
yes great but what if the III pair is not chosen,( say 2 identical one's are chosen from LL and SS) then we have only 2 choices from paired letters and remaining 4 that gives us (4+2) ...this will create a different answer .....
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by AndyB » Tue Mar 15, 2011 7:40 am
Hi Mitch,

You are absolutely right.Thanks for solving the confusion.

The real solution of the problem lies in understanding the term..." 2 alike, the other 2 distinct"

Yes the confusion I had was with usage of I,I,I.But when we use I,I as the like pair the remaining I cannot be used which then converts the problem to "3 alike , one different" problem.

Guys the OA is 84

Thanks once again,
AndyB.
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