What is the value of integer n ? (1) n(n + 1) = 20 (2) 6!/(n!)(6-n)!

Question

What is the value of integer n ? (1) n(n + 1) = 20 (2) 6!/(n!)(6-n)! = 15 Source: www.GMATinsight.com

vasuca10 · Answer

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!*(6-n)! = 15
720/n!*(6-n)! = 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

GMATinsight · Answer

vasuca10  the highlighted part of your explanation is flawed. n(n+1) refers to the product of two consecutive integers

vasuca10 · Answer

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

GmatDaddy · Answer

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.

fskilnik · Answer

Obs.: 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 ight)\,\,\,20 = \left\{ \begin{gathered}
 \,\,4 \cdot 5\,\,\, \Rightarrow \,\,\,n\, = 4\,\, \hfill \
 \left( { - 5} ight) \cdot \left( { - 4} ight)\,\,\, \Rightarrow \,\,\,n = \, - 5\,\,\, \hfill \ 
\end{gathered} ight.\,\,\,\, \Rightarrow \,\,\,\,\,\,{	ext{INSUF}}.\,\,

Important: we found TWO distinct real roots for the SECOND-degree equation given in statement (1). 
Conclusion: there are NO other possibilities for n (even when nonintegers are considered, by the way!)

\left( 2 ight)\,\,15 = \frac{{6!}}{{n!\,\,\left( {6 - n} ight)!}} = C\left( {6,n} ight)\,\,\,\, \Rightarrow \,\,\,n = 2\,\,\,{	ext{or}}\,\,\,n = 4\,\,\,\, \Rightarrow \,\,\,\,{	ext{INSUF}}.\,\,\,

\left( {1 + 2} ight)\,\,\,\,\,\,\,n = 4\,\,\,\, \Rightarrow \,\,\,\,{	ext{SUF}}.

This solution follows the notations and rationale taught in the GMATH method.

Regards,
fskilnik.

MattyE · Answer

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

fskilnik · Answer

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(6,6-n), 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=n-p, therefore C(n,p) = C(n,n-p) where these notations make sense, of course!

I hope you understood (and liked) my comments.

Regards,
Fabio.

GMATinsight · Answer

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(n-r) I hope this helps!!!

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What is the value of integer n ? (1) n(n + 1) = 20 (2) 6!/(n!)(6-n)!

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