4

I am attempting to solve an equation ${{n-2} \choose {2}} + {{n-3} \choose {2}} + {{n-4} \choose {2}} = 136$. With the formula for a combination being $\frac{n!}{r!(n - r)!}$, I simplified the given equation to:

$(n-2)! + (n-3)! + (n-4)! = 272n - 1088$

However, I am not sure how I would solve for $\\n$ in the simplified version. I have looked into Sterling's Approximation, hoping to find some way to solve this there, but I believe that is going about it incorrectly.

Any help that can be offered in solving this equation is greatly appreciated.

nmagerko
  • 505

4 Answers4

5

Note that $$\dbinom{n-2}2 = \dfrac{(n-2)(n-3)}{2}$$ $$\dbinom{n-3}2 = \dfrac{(n-3)(n-4)}{2}$$ $$\dbinom{n-4}2 = \dfrac{(n-4)(n-5)}{2}$$ Hence, $$\dbinom{n-2}2 + \dbinom{n-3}2 + \dbinom{n-4}2 = \dfrac{(n-2)(n-3)}{2} + \dfrac{(n-3)(n-4)}{2} + \dfrac{(n-4)(n-5)}{2}$$ Now solve the quadratic to get the answer.

EDIT $$\dbinom{m}2 = \dfrac{m!}{2! (m-2)!} = \dfrac{m(m-1) \cdot (m-2)!}{2! \cdot (m-2)!} = \dfrac{m(m-1)}2$$

1

It is best to leave factorials out of this: $\dbinom{x}{2}=\dfrac{x(x-1)}{2}$.

Remarks: $1.$ The binomial coefficients were simplified incorrectly.

$2.$ We don't even need "algebra" to solve the problem. Evaluate your expression, for $n=4, 5, \dots$. After a short while your sum is clearly too big.

$3.$ We describe a way to calculate a binomial coefficient, say $\dbinom{10}{3}$. By the formula you used, this is $\dfrac{10!}{3!7!}$.

Look at the top. This is $(10)(9)(8)(7)(6)(5)(4)(3)(2)(1)$. Note that it is equal to $(10)(9)(8)(7!)$. Now the $7!$ part cancels the $7!$ at the bottom.

So our binomial coefficient is equal to $\dfrac{(10)(9)(8)}{3!}$. We did not have to calculate a big factorial.

For something like $\dbinom{x}{2}$, use $\dfrac{x!}{2!(x-2)!}$. The top is $x(x-1)(x-2)!$. The $(x-2)!$ cancels the one at the bottom, and we end up with $\dfrac{x(x-1)}{2!}$.

André Nicolas
  • 507,029
  • I think my ignorance in this is understated... I do not understand how to evaluate binomial coefficients. I have never heard of them before today! – nmagerko Feb 23 '13 at 19:54
  • 1
    Maybe earlier in your school work, what we call $\binom{n}{r}$ was called $C(n,r)$, or $C_r^n$, or something like that. (There are many "school" notations for the binomial coefficients.) Mathematicians almost always use $\binom{n}{r}$. – André Nicolas Feb 23 '13 at 20:06
  • Yes, I recall that notation being equivalent to $C(n,r)$. I believe I understand what you were saying now - thank you very much! – nmagerko Feb 23 '13 at 20:07
1

Bad simplification, that one. $$ \begin{align*} \binom{n - 2}{2} + \binom{n - 3}{2} + \binom{n - 4}{2} &= \frac{(n - 2)(n -3)}{2} + \frac{(n - 3) (n - 4)}{2} + \frac{(n - 4)(n - 5)}{2} \\ &= \frac{3 n^2 - 21 n + 38}{2} \end{align*} $$ The binomial coefficient $\binom{n}{k} = \binom{n}{n - k} = \frac{n (n - 1) \ldots (n - k + 1)}{k!}$ ($k$ factors upstairs and downstairs) in ubicuous in combinatorics.

You surely can take it from there...

vonbrand
  • 27,812
  • I understand that my original approach was too algebraic to work, but I have never heard of binomial coefficients. Any explanation you can give within your answer would be very helpful. – nmagerko Feb 23 '13 at 19:56
  • 2
    @nmagerko, no, your approach was fine, just that the expansion went too far. If plan A doesn't work, try plan B... don't get stuck on one approach. – vonbrand Feb 23 '13 at 20:03
0

$n=13$ after solving this equation $3n^2-21n-234=0 $

rurouniwallace
  • 6,266
Ramez Hindi
  • 581
  • Hello Ramez Hindi, welcome to math SE. While your answer is an accurate statement, it would be helpful if you went into more detail as to how it relates to the problem stated. Also, see our guide to typing mathjax to format your mathematics in a more legible format. – rurouniwallace Aug 05 '13 at 04:17