Given a set X, if the subsets of X with exactly 3 elements is 14 more than the subsets of X with exactly 2 elements; then how many subsets of X have exactly 4 elements?

Question

Given a set X,  if the subsets of X with exactly 3 elements is 14 more than the subsets of X with exactly 2 elements; then how many subsets of X have exactly 4 elements?

amistre64 · Answer

im thinking its 35, but thats just me :)

parthkohli · Answer

Is this possible? I mean, the number of subsets with 2 elements should ideally be greater than the number of subsets with 3 elements, right?

Or did I misinterpret the meaning of the question?

amistre64 · Answer

you might be misreading the question

amistre64 · Answer

think of it like this; what is the number of way of choosing 3 elements from a set of "n", if permuations is not acceptable?

parthkohli · Answer

$$\binom{n}{3}$$

terenzreignz · Answer

Fancy ... 35 seems to be correct...
But I found no other way to get it except trial and error...
There must be a more efficient way :(

amistre64 · Answer

and (n 3) = (n 2) + 14 , if n=?

terenzreignz · Answer

n = 7

amistre64 · Answer

then (7 4) = 35

hartnn · Answer

i too get n=7

parthkohli · Answer

Ah!

terenzreignz · Answer

How did you get n = 7 guys? T.T

hartnn · Answer

(n choose 2)+14 = (n choose 3)

hartnn · Answer

so, yes 7 choose 4 = 35 subsets

amistre64 · Answer

$$\frac{n(n-1)(n-2)}{3!}=\frac{n(n-1)}{2!}+14$$
$$n(n-1)(n-2)=3n(n-1)+14(6)$$

terenzreignz · Answer

oh... that simplifies things :D
I must have panicked, @amistre64  
It was... a momentary madness~

amistre64 · Answer

or, you could have pascaled a triangle :)

terenzreignz · Answer

It's a perfect cube?

amistre64 · Answer

not sure if the n stuff above is perfect cubish, but it does have 7 as one of the possible solutions to it; and the only solution that made sense

terenzreignz · Answer

I'd have just simplified the thing, and then use the rational root test, after all, it's looking for an integer solution.

Anyway, I did trial and error'ed it, but that's definitely not the best way to go about it...

I learned :D

hartnn · Answer

n= 7 is only integer solution,
 other 2 are complex

amistre64 · Answer

lol, yeah, i just did a wolf run and 7 is the only integer soluton :)

amistre64 · Answer

n(n−1)(n−2)=3n(n−1)+14(6)

n^3-3n^2+2n = 3n^2-3n+84

n^3-6n^2+5n - 84 = 0

i think i got to here before going to the pascal

terenzreignz · Answer

Rational root test... at least it limits your integer solutions :D

amistre64 · Answer

yep, and cardano would be fine as well; let n = (u+6/3)

(u+2)^3 - 6(u+2)^2 + 5(u+2) - 84 = 0

u^3 +6u^2 +12u +8
        -6u^2 -24u -24
                   +5u +10
                           -84
-------------------
u^3             -7u - 90 = 0

amistre64 · Answer

rational roots would at least be simpler :)  

1,84
2,42
4,21

3,28
14,6

5,x

7,12

amistre64 · Answer

lol, thatd be more readable as 

3,28
6,14
12,7

double the left, half the right

hartnn · Answer

hmm...doing manually is quite cumbersome.