product of 3 numbers from a set of 7 numbers

Question

Set $S$ consists of the integers from -1 to 5, inclusive. If $N$ is the product of three distinct members of Set $S$, how many unique values of $N$ are there?

I thought of it like this--> there are $7$ numbers so if we multiply by any $3$ out of it , number of unique values of $N$ will be $^7C_3$ ..because of $0$ it will be reduced though ..coz it will be repeated...so in $^6C_2$ it will be repeated...so answer is $^7C_3 - \space ^6C_2 = 35-15=20..$

Please let me know

Lord Soth · Answer 1 · 2013-11-05T20:01:48.063

2

You have thrown out all the zeroes! Add the zero and you are done.

As a precautionary note, your argument will not hold for an arbitrary set $S$. For example if $S$ were $S = \{-2,-1,\ldots,1,2,3,4\}$, then $-2 \times -1 \times 4 = 1\times 2 \times 4 = 8$, so we would have counted $8$ twice. I assume you made sure that the products that do not contain a zero will be all distinct for your $S$.

edited Nov 05 '13 at 20:01

answered Nov 05 '13 at 19:56

Lord Soth

7,750
20
37

Thanks Lord soth..... and very thanks for pointing out the effect of 2 negative numbers in the set... :) – atir Nov 05 '13 at 20:18
@atir You are welcome. – Lord Soth Nov 05 '13 at 20:22

score 0 · Answer 2 · answered Nov 05 '13 at 20:30

Yes, you're missing only the zero.

To help you check your solution, I've used this Perl code:

for $i (-1..5) {
  for $j (-1..5) {
    next unless $i-$j;
    for $k (-1..5) {
      next unless $i-$k && $j-$k;
      $x{$i*$j*$k} = 1;
    }
  }
}
print scalar(keys %x), " distinct products: ", join ", ", sort {$a<=>$b} keys %x;

and it provided the following solution:

21 distinct products: -20, -15, -12, -10, -8, -6, -5, -4, -3, -2, 0, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60

product of 3 numbers from a set of 7 numbers

2 Answers2