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What is the value of integer n ? (1) n(n + 1) = 20 (2) 6!/(n!)(6n)!
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08 Sep 2018, 04:24
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What is the value of integer n ? (1) n(n + 1) = 20 (2) 6!/(n!)(6n)! = 15 Source: www.GMATinsight.com
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Re: What is the value of integer n ? (1) n(n + 1) = 20 (2) 6!/(n!)(6n)!
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08 Sep 2018, 04:41
Option C As per Statement 1 n*(n+1)=20 Possible values 1*20 , 1*20, 2*10, 2*10, 4*5, 4*5, 5*4, 5*4, 10*2, 10*2 Multiple values hence Statement 1 alone is not suffice As per Statement 2 6!/n!*(6n)! = 15 720/n!*(6n)! = 15 which implies n can be 4 as well as 2 Hence statement 2 alone itself is not suffice Now when we combine both statements all negative values which we deduced from statement 1 can be eliminated as factorial can never be negative. Values satisfying both the statements is n =4 Hence Option C Kindly give kudos if my explanation helped
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Re: What is the value of integer n ? (1) n(n + 1) = 20 (2) 6!/(n!)(6n)!
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08 Sep 2018, 04:44
vasuca10 wrote: Option C As per Statement 1 n*(n+1)=20 Possible values 1*20 , 1*20, 2*10, 2*10,4*5, 4*5, 5*4, 5*4, 10*2, 10*2Multiple values hence Statement 1 alone is not suffice As per Statement 2 6!/n!*(6n)! = 15 720/n!*(6n)! = 15 which implies n can be 4 as well as 2 Hence statement 2 alone itself is not suffice Now when we combine both statements all negative values which we deduced from statement 1 can be eliminated as factorial can never be negative. Values satisfying both the statements is n =4 Hence Option C Kindly give kudos if my explanation helped vasuca10 the highlighted part of your explanation is flawed. n(n+1) refers to the product of two consecutive integers
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Re: What is the value of integer n ? (1) n(n + 1) = 20 (2) 6!/(n!)(6n)!
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08 Sep 2018, 04:47
GMATinsight wrote: vasuca10 wrote: Option C As per Statement 1 n*(n+1)=20 Possible values 1*20 , 1*20, 2*10, 2*10,4*5, 4*5, 5*4, 5*4, 10*2, 10*2Multiple values hence Statement 1 alone is not suffice As per Statement 2 6!/n!*(6n)! = 15 720/n!*(6n)! = 15 which implies n can be 4 as well as 2 Hence statement 2 alone itself is not suffice Now when we combine both statements all negative values which we deduced from statement 1 can be eliminated as factorial can never be negative. Values satisfying both the statements is n =4 Hence Option C Kindly give kudos if my explanation helped vasuca10 the highlighted part of your explanation is flawed. n(n+1) refers to the product of two consecutive integers Oh ohk got it the possible values as per statement 1 shall be 4,5 and 5 and 4 I got my mistake thanks for rectifying I solved it in a bit of hurry
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Re: What is the value of integer n ? (1) n(n + 1) = 20 (2) 6!/(n!)(6n)!
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08 Sep 2018, 05:04
vasuca10 wrote: Option C As per Statement 1 n*(n+1)=20 Possible values 1*20 , 1*20, 2*10, 2*10, 4*5, 4*5, 5*4, 5*4, 10*2, 10*2 Multiple values hence Statement 1 alone is not suffice Now when we combine both statements all negative values which we deduced from statement 1 can be eliminated as factorial can never be negative. Values satisfying both the statements is n =4 Hence Option C Kindly give kudos if my explanation helped For your evaluation of statement 2, only 2 cases will exist. Since it is already given in question stem that n is an integer, therefore either (4,5) or(5,4) as n*(n+1) means the product of two consecutive integers. and yes when you combine it with Statement 2, it will yield the answer C.
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What is the value of integer n ? (1) n(n + 1) = 20 (2) 6!/(n!)(6n)!
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08 Sep 2018, 15:16
GMATinsight wrote: What is the value of integer n ? (1) n(n + 1) = 20 (2) 6!/ [(n!)(6n)! ] = 15 Source: http://www.GMATinsight.comObs.: in GMAT`s universe, factorials are defined ONLY for nonnegative integer values. Therefore (when considering) statement (2) we MUST implicitly ASSUME that n is equal to 0,1,2,3,4,5 or 6. \(\left( 1 \right)\,\,\,20 = \left\{ \begin{gathered} \,\,4 \cdot 5\,\,\, \Rightarrow \,\,\,n\, = 4\,\, \hfill \\ \left( {  5} \right) \cdot \left( {  4} \right)\,\,\, \Rightarrow \,\,\,n = \,  5\,\,\, \hfill \\ \end{gathered} \right.\,\,\,\, \Rightarrow \,\,\,\,\,\,{\text{INSUF}}.\,\,\) Important: we found TWO distinct real roots for the SECONDdegree equation given in statement (1). Conclusion: there are NO other possibilities for n (even when nonintegers are considered, by the way!) \(\left( 2 \right)\,\,15 = \frac{{6!}}{{n!\,\,\left( {6  n} \right)!}} = C\left( {6,n} \right)\,\,\,\, \Rightarrow \,\,\,n = 2\,\,\,{\text{or}}\,\,\,n = 4\,\,\,\, \Rightarrow \,\,\,\,{\text{INSUF}}.\,\,\,\) \(\left( {1 + 2} \right)\,\,\,\,\,\,\,n = 4\,\,\,\, \Rightarrow \,\,\,\,{\text{SUF}}.\) This solution follows the notations and rationale taught in the GMATH method. Regards, fskilnik.
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Re: What is the value of integer n ? (1) n(n + 1) = 20 (2) 6!/(n!)(6n)!
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17 Sep 2018, 22:13
Hey all,
I didn't spot that #2 can have both 4 and 2 as a solution.
Is there a way to quickly/easily identify that a variable within combination has more than 1 solution?
Thanks in advance, Matt



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Re: What is the value of integer n ? (1) n(n + 1) = 20 (2) 6!/(n!)(6n)!
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18 Sep 2018, 02:50
MattyE wrote: Hey all,
I didn't spot that #2 can have both 4 and 2 as a solution.
Is there a way to quickly/easily identify that a variable within combination has more than 1 solution?
Thanks in advance, Matt Hi, Matt! Nice question! YES, if you "imagine" C(6,n) counts the number of groups of n people among 6 people, you will probably remember two things: 1. You expect n to be only 0, 1, 2, 3, 4, 5 or 6. (Remember that 0!=1 by definition, hence C(6,0) = 6! divided by 0! times 6!, hence C(6,0)=1.) 2. You will remember that C(6,n) equals C(6,6n), example, C(6,2) = C(6,4)... reason: When you choose 2 people (for the group), you automatically choose 4 people (NOT for the group)... in other words, If p+k = n, then C(n,p) = C(n,k) where k=np, therefore C(n,p) = C(n,np) where these notations make sense, of course! I hope you understood (and liked) my comments. Regards, Fabio.
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What is the value of integer n ? (1) n(n + 1) = 20 (2) 6!/(n!)(6n)!
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18 Sep 2018, 18:37
MattyE wrote: Hey all,
I didn't spot that #2 can have both 4 and 2 as a solution.
Is there a way to quickly/easily identify that a variable within combination has more than 1 solution?
Thanks in advance, Matt MattyE There is always more than one solutions Look at it this way, if you select 2 out of 5 objects then you leave 3 which is equivalent to selecting 3 out of 5 hence 5C2 = 5C3 Similarly, if you select 3 out of 7 objects then you leave 4 which is equivalent to selecting 4 out of 7 hence 7C3 = 7C4 i.e. nCr = nC(nr)I hope this helps!!!
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What is the value of integer n ? (1) n(n + 1) = 20 (2) 6!/(n!)(6n)! &nbs
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