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To furnish a room in a model home, an interior decorator is [#permalink]
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To furnish a room in model home, an interior decorator is to select 2 chairs and 2 tables from a collection of chairs and tables in a warehouse that are all different from each other. If there are 5 chairs in the warehouse and if 150 different combinations are possible, how many tables are in the warehouse? A. 6 B. 8 C. 10 D. 15 E. 30 The answer is A. BUT! what is the next step in the process?
5C2*nC2=150 10*nC2=150 nC2=15
n! / 2!*(n2)! = 15 n!=30(n2)!
stuck here. what do I need to do next in order to arrive to the answer choice A, 6 tables?
award kudos for good explanation
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barakhaiev wrote: HI!
This question has already been posted several times, but still I did not understand the reasoning:
To furnish a room in a model home, an interior decorator is to select 2 chairs from a collection of chairs and tables in a warehouse that are all different from each other. If there are 5 chairs in the warehouse and if 150 different combinations are possible, how many tables are in the warehouse?
6 8 10 15 30
The answer is A. BUT! what is the next step in the process?
5C2*nC2=150 10*nC2=150 nC2=15
n! / 2!*(n2)! = 15 n!=30(n2)!
stuck here. what do I need to do next in order to arrive to the answer choice A, 6 tables?
award kudos for good explanation You've done most of the work right, next step: \(\frac{n!}{2!*(n2)!}=15\). \((n2)!\) will just cancel out and will get: \(\frac{(n1)*n}{2!}=15\) > \((n1)*n=30\) > \(n=6\) Hope it's clear.
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01 Nov 2009, 10:20
Thank you very much, Bunuel, for such a fast and detailed response. Kudos, as promised!
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23 Sep 2010, 12:04
udaymathapati wrote: Pls help to solve this problem. Attachment: Image3.JPG No of combinations = C(5,2) * C(x,2) = 10 * x(x1)/2 =150 Where x is the number of tables So x(x1)/2 = 15 x(x1) = 30 And we know x is an integer => x=6
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27 Sep 2010, 07:38
Shrouded how did you arrive at x(x1)/2 for c(x,2)? shouldn't this be x!/2!(x2)! ? I am having trouble following how you arrived at x(x1)/2? Please advise, thank you!



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27 Sep 2010, 08:13
Out of curiosity Why didn't we take different possible arrangements where order of selection is important (permutation). For example, if we have two chairs (A&B) and two tables (X&Y) selected from the warehouse, we can arrange them as AX & BY and AY & BX. Was this because of the fact that it was given that there are 150 different combinations ?
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27 Sep 2010, 08:16
gettinit wrote: Shrouded how did you arrive at x(x1)/2 for c(x,2)? shouldn't this be x!/2!(x2)! ? I am having trouble following how you arrived at x(x1)/2? Please advise, thank you! x ! / [2!(x2)!] = [(x2)!(x1)(x)]/[2!(x2)!] = [(x1)(x)]/2
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27 Sep 2010, 08:31
adishail wrote: Out of curiosity
Why didn't we take different possible arrangements where order of selection is important (permutation). For example, if we have two chairs (A&B) and two tables (X&Y) selected from the warehouse, we can arrange them as AX & BY and AY & BX. Was this because of the fact that it was given that there are 150 different combinations ? Decorator is to select 2 different chairs out of 5 and 2 different tables out of n. Now, two selections {C1, C2, T1, T2} and {C2, C1, T2, T1,} are the same for decorator as there are the same chairs and tables in both of them. So the order in which the items are taken is not considered. Hope it's clear.
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27 Sep 2010, 08:54
Bunuel wrote: adishail wrote: Out of curiosity
Why didn't we take different possible arrangements where order of selection is important (permutation). For example, if we have two chairs (A&B) and two tables (X&Y) selected from the warehouse, we can arrange them as AX & BY and AY & BX. Was this because of the fact that it was given that there are 150 different combinations ? Decorator is to select 2 different chairs out of 5 and 2 different tables out of n. Now, two selections {C1, C2, T1, T2} and {C2, C1, T2, T1,} are the same for decorator as there are the same chairs and tables in both of them. So the order in which the items are taken is not considered. Hope it's clear. Yes. I understood your explanation. However, I am talking about a case where we would have {C1T1, C2T2, and C1T2, C2T1} since these two can be considered different arrangements. But I think I somehow got stuck with the idea of making a set of a chair and table. I guess if they would have worded it something like  If 150 different sets of tables and chair are possible ...........then order would have been important. Is that correct Bunuel ? And in that case we would have taken COMBINATION for chairs and PERMUTATION for tables or vice versa because if we would have taken PERMUTATION for both tables an chairs, then we would have ended up with two identical cases like {C1T1, C2T2} and {C2T2, C1T1}.
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27 Sep 2010, 09:23
adishail wrote: Bunuel wrote: adishail wrote: Out of curiosity
Why didn't we take different possible arrangements where order of selection is important (permutation). For example, if we have two chairs (A&B) and two tables (X&Y) selected from the warehouse, we can arrange them as AX & BY and AY & BX. Was this because of the fact that it was given that there are 150 different combinations ? Decorator is to select 2 different chairs out of 5 and 2 different tables out of n. Now, two selections {C1, C2, T1, T2} and {C2, C1, T2, T1,} are the same for decorator as there are the same chairs and tables in both of them. So the order in which the items are taken is not considered. Hope it's clear. Yes. I understood your explanation. However, I am talking about a case where we would have {C1T1, C2T2, and C1T2, C2T1} since these two can be considered different arrangements. But I think I somehow got stuck with the idea of making a set of a chair and table. I guess if they would have worded it something like  If 150 different sets of tables and chair are possible ...........then order would have been important. Is that correct Bunuel ? And in that case we would have taken COMBINATION for chairs and PERMUTATION for tables or vice versa because if we would have taken PERMUTATION for both tables an chairs, then we would have ended up with two identical cases like {C1T1, C2T2} and {C2T2, C1T1}. You are right we are not told to pair a chair with table. Also I don't think that changing the word "combination" with "set" would make a difference, the question is clearly talking about the different selections of 2 chairs and 2 tables. If we were interested in such pairs then you should notice that one particular selection of 2 chairs and 2 tables {C1, C2, T1, T2} can give 2 different pairs of chair/table: {C1/T1, C2/T2} or {C1/T2, C2/T1}.
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27 Sep 2010, 09:35
Quote: If we were interested in such pairs then you should notice that one particular selection of 2 chairs and 2 tables {C1, C2, T1, T2} can give 2 different pairs of chair/table: {C1/T1, C2/T2} or {C1/T2, C2/T1}. Correct. In that case, either the total combinations can be 1) multiplied by 2, or 2) find Permutation for both items and then divide by 2 to eliminate the identical set like {C1,T1 & C2,T2} and {C2,T2 & C1,T1} Is that right ?
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barakhaiev wrote: HI!
This question has already been posted several times, but still I did not understand the reasoning:
To furnish a room in a model home, an interior decorator is to select 2 chairs Combo box arrangement for chairs (_)(_)/2! barakhaiev wrote: and 2 tables from a colelction of chairs and tables in a warehouse that are all different from each other. Combo box arrangement for chairs and tables (_)(_)/2! * (_)(_)/2! barakhaiev wrote: If there are 5 chairs in the warehouse Fill in combo box arrangement (5)(4)/2! * (_)(_)/2! barakhaiev wrote: and if 150 different combinations are possible (5)(4)/2! * (_)(_)/2! = 150 barakhaiev wrote: , how many tables in the warehouse? (5)(4)/2! * (Tables)(Tables  1)/2! = 150 Algebra (Tables)(Tables  1) = 150 * 2! * 2 ! / (5)(4) = 30 = 6*5 Tables = 6



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GMAT prep problem... It should be solved like following Suppose we have to select 2 tables from “X” tables We have already given that we have to select 2 chairs form 5 and we total possible combinations are 150
Therefore, 5C2 * XC2 = 150 5*4/2 * X!/2!(X2)! = 150 X!/2!(X2)! = 15 X!/(X2)! = 30 X(X1)(X2)! / (X2)! = 30 X(X1) = 30 X2 – X – 30 = 0 X2 – 6X + 5X – 30 = 0 X(X – 6) + 5(X – 6) = 0 (X – 6) (X + 5) = 0 X = 6 or X = 5 Negative value is not possible since we are calculating the number of tables Therefore X = 6 Hence “A” is the answer



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Updated on: 26 Jul 2015, 17:41
Bunuel wrote: gettinit wrote: Shrouded how did you arrive at x(x1)/2 for c(x,2)? shouldn't this be x!/2!(x2)! ? I am having trouble following how you arrived at x(x1)/2? Please advise, thank you! \(C(x,2)=\frac{x!}{2!*(x2)!}=\frac{(x2)!*(x1)*x}{2!*(x2)!}\) so you can reduce by \((x2)!\) and you'll get \(\frac{(x1)x}{2}\). Hi Bunuel, I don't understand how you got (x2)!*(x1)*x in the numerator
Originally posted by Rookie124 on 26 Jul 2015, 17:39.
Last edited by Rookie124 on 26 Jul 2015, 17:41, edited 1 time in total.



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To furnish a room in a model home, an interior decorator is [#permalink]
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26 Jul 2015, 18:00
mawus wrote: Bunuel wrote: gettinit wrote: Shrouded how did you arrive at x(x1)/2 for c(x,2)? shouldn't this be x!/2!(x2)! ? I am having trouble following how you arrived at x(x1)/2? Please advise, thank you! \(C(x,2)=\frac{x!}{2!*(x2)!}=\frac{(x2)!*(x1)*x}{2!*(x2)!}\) so you can reduce by \((x2)!\) and you'll get \(\frac{(x1)x}{2}\). Hi Bunuel, I don't understand how you got (x2)!*(x1)*x in the numerator Let me try to answer this question. When you write: \(C(x,y) = \frac{x!}{y!*(xy)!}\) Now substitute y =2 in the above equation: \(C(x,2) = \frac{x!}{2!*(x2)!}\) Now any factorial, n! = n*(n1)*(n2).....1 Thus, x! = x*(x1)*[(x2)*(x3).....1] and particularly, [(x2)*(x3).....1] = (x2)! ( we are doing this to cancel out (x2)! in the denominator. Finally, \(C(x,2) = \frac{x!}{2!*(x2)!}\) > \(C(x,2) = \frac{x*(x1)*(x2)!}{2!*(x2)!}\) If these variables are creating confusion for you, take 4! = 4*3*2*1 > If x= 4, then x1 =3, x2 =2, ...etc. This is what Bunuel has done above. If you are still having difficulties, for the above question, 150 = C(5,2) * C(n,2), where n= total number of tables in the warehouse. \(C(5,2) = \frac{5!}{2!*3!} = \frac{5*4*3!}{2!*3!}\) = 10 Thus, C(n,2) = 150/10 = 15 Now, try the given answer choices to see which of the values when substituted for n gives you C(n,2) =15. n =6 will satisfy this equation. Hope this helps.



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Re: To furnish a room in a model home, an interior decorator is [#permalink]
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18 Sep 2017, 05:50
barakhaiev wrote: To furnish a room in model home, an interior decorator is to select 2 chairs and 2 tables from a collection of chairs and tables in a warehouse that are all different from each other. If there are 5 chairs in the warehouse and if 150 different combinations are possible, how many tables are in the warehouse?
A. 6 B. 8 C. 10 D. 15 E. 30 We are given that an interior decorator is to select 2 chairs and 2 tables from a collection of chairs and tables. We are also given that there are 5 chairs in the warehouse and 150 different possible combinations. We must determine the number of tables. We can let n = the number of tables and create the following equation: 5C2 x nC2 = 150 [(5 x 4)/2!] x [(n x n1)/2!] = 150 20/2 x (n^2 – 1)/2 = 150 10 x (n^2 – 1)/2 = 150 (n^2 – 1)/2 = 15 n^2 – 1 = 30 n^2 – 1 – 30 = 0 (n – 6)(n + 5) = 0 n = 6 or n = 5. Since n must be positive, the number of tables is 6. Answer: A
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Re: To furnish a room in a model home, an interior decorator is [#permalink]
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14 Feb 2018, 08:05
Bunuel wrote: barakhaiev wrote: HI!
This question has already been posted several times, but still I did not understand the reasoning:
To furnish a room in a model home, an interior decorator is to select 2 chairs from a collection of chairs and tables in a warehouse that are all different from each other. If there are 5 chairs in the warehouse and if 150 different combinations are possible, how many tables are in the warehouse?
6 8 10 15 30
The answer is A. BUT! what is the next step in the process?
5C2*nC2=150 10*nC2=150 nC2=15
n! / 2!*(n2)! = 15 n!=30(n2)!
stuck here. what do I need to do next in order to arrive to the answer choice A, 6 tables?
award kudos for good explanation You've done most of the work right, next step: \(\frac{n!}{2!*(n2)!}=15\). \((n2)!\) will just cancel out and will get: \(\frac{(n1)*n}{2!}=15\) > \((n1)*n=30\) > \(n=6\) Hope it's clear. Can u pls explain why we are doing this? n! / 2!*(n2)! = 15




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