How many people bought TVs at the Electronics Depot?

Question

(1) One out of every six Electronics Depot customers bought a TV.
(2) If four Electronics Depot customers are chosen at random there are 126 different groups of people that could be chosen.

rajudantuluri · Answer

Statement 1) 1 in 6 bought a TV. We don't know how many customers so NOT SUFFICIENT.

Statement 2) Picking 4 people for a group out of a certain number of people would result in 126 combinations. Let's say there are K customers. K!/4!*(K-1)!=126. Deducing this gives us K=9. So we know number of Customers but we do not know how many bought TV.

Statement 1) Gives ratio of number of people who bought TV to Customers and Statement 2) Gives the number of customers.

So C, Both required but neither alone sufficient.

paolodeppa · Answer

Statement 2) Picking 4 people for a group out of a certain number of people would result in 126 combinations. Let's say there are K customers. K!/4!*(K-1)!=126. Deducing this gives us K=9. So we know number of Customers but we do not know how many bought TV.

Are you sure about the formula you used to solve Statement 2? shouldn't it be K!/4!*(K -4 )!=126??

fskilnik · Answer

? = P\,\,\,\,\left( {{\rm{\# }}\,\,{\rm{bought - TV - there}}\,\,{\rm{people}}} \right)

\left( 1 \right)\,\,P = {N \over 6}\,\,\,\left\{ \matrix{
 \,{\rm{Take}}\,\,N = 6\,\,{\rm{customers}}\,\,\,\, \Rightarrow \,\,\,? = P = 1 \hfill \cr 
 \,{\rm{Take}}\,\,N = 12\,\,{\rm{customers}}\,\,\,\, \Rightarrow \,\,\,? = P = 2 \hfill \cr} \right.\,\,\,\,\,\,\,\,\,\,\,\left( {N = \# \,\,{\rm{customers}}\,\,{\rm{there}}} \right)

\left( 2 \right)\,\,C\left( {N,4} \right) = 126\,\,\,\,\, \Rightarrow \,\,\,\,\,N\,\,{\rm{unique}}\,\,\,\left( {{\rm{and}}\,\,N > 4\,\,{\rm{for}}\,\,{\rm{sure}}} \right)\,\,\,\,\,\left\{ \matrix{
 \,{\rm{Take}}\,\,P = 1\,\,\,\, \Rightarrow \,\,\,? = 1 \hfill \cr 
 \,{\rm{Take}}\,\,P = 2\,\,\,\, \Rightarrow \,\,\,? = 2 \hfill \cr} \right.

\left( {1 + 2} \right)\,\,{\rm{unique}}\,\,N\,\, = 6P\,\,\,\,\, \Rightarrow \,\,\,\,\,P\,\,{\rm{unique}}

The correct answer is therefore (C), indeed.

We follow the notations and rationale taught in the  GMATH  method.

Regards, 
Fabio.

Fdambro294 · Answer

Is it possible to get a unique answer?

Together:

1 out of every 6 customers bought a TV

And

If we are making unique groups of 4 people —- there must be 9 total customers from which we select the groups of 4 people

9! / (4! * 5!) = 9 * 8 * 7 * 6 / 4 * 3 * 2 = 9 * 2 * 7 = (9) (14) =

126 different groups

Since we can’t have a (1/2) of a person:

How can it be that 1 out of every 6 customers bought a TV, but there are 9 total customers?

Posted from my mobile device

anuhyadixit · Answer

Can someone give a clear explanation for this?

Bunuel   KarishmaB

KarishmaB · Answer

Yes, we do get a unique answer but it is a fraction for number of people! That's not ok.

If 4 random people are selected, we can make 126 distinct groups. This means that if number of customers is n, 
nC4 = 126
There is only one value of n for which this will hold and that is 9 (with some hit & trial). Note that nC4 is unique with every different value of n (n is an integer >= 4). 
5C4 = 5
6C4 = 15
... and so on... nC4 will keep increasing as n keeps increasing. 
So since 9C4 = 126, n must be 9 only.

But if every sixth person buys a TV, we get 1.5 people of the 9 bought a TV. That's odd!

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How many people bought TVs at the Electronics Depot?

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