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6. If [tex][tex]$12 \binom{r}{2} = \binom{r+2}{3}$[/tex][/tex], find [tex][tex]$r$[/tex][/tex].

[Hint: [tex] 12 \cdot \frac{r(r-1)}{2 \times 1} = \frac{(r+2)(r+1)r}{3 \times 2 \times 1} \Rightarrow 18 \cdot (r-1) = (r+1) \cdot 2(r-1)[/tex]]

Sagot :

GinnyAnswer
To solve the equation given in the hint, we need to find the value of [tex]\( r \)[/tex] for which the equation holds true. We start with the combinatorial notation and the given hint. The given equation is:

[tex]\[ 12 \cdot {^r}C_2 = {^{r+2}}C_3 \][/tex]

We know the general formula for combinations, where [tex]\({^n}C_k = \frac{n!}{k!(n-k)!} \)[/tex]. Applying this formula, we get:

[tex]\[ {^r}C_2 = \frac{r!}{2!(r-2)!} = \frac{r(r-1)}{2 \times 1} = \frac{r(r-1)}{2} \][/tex]

[tex]\[ {^{r+2}}C_3 = \frac{(r+2)!}{3!(r-3)!} = \frac{(r+2)(r+1)r}{3 \times 2 \times 1} = \frac{(r+2)(r+1)r}{6} \][/tex]

Plugging these back into the given equation:

[tex]\[ 12 \cdot \frac{r(r-1)}{2} = \frac{(r+2)(r+1)r}{6} \][/tex]

Now simplify each side:

[tex]\[ 12 \cdot \frac{r(r-1)}{2} = 6r(r-1) \][/tex]

So, the equation becomes:

[tex]\[ 6r(r-1) = \frac{(r+2)(r+1)r}{6} \][/tex]

Multiplying both sides by 6 to clear the fraction, we get:

[tex]\[ 36r(r-1) = (r+2)(r+1)r \][/tex]

Dividing both sides by [tex]\( r \)[/tex] (assuming [tex]\( r \neq 0 \)[/tex]):

[tex]\[ 36(r-1) = (r+2)(r+1) \][/tex]

Expanding the right side:

[tex]\[ 36(r-1) = r^2 + 3r + 2 \][/tex]

Distribute and simplify the left side:

[tex]\[ 36r - 36 = r^2 + 3r + 2 \][/tex]

Bringing all terms to one side to set the equation to zero:

[tex]\[ r^2 + 3r + 2 - 36r + 36 = 0 \][/tex]

[tex]\[ r^2 - 33r + 38 = 0 \][/tex]

Now, we solve this quadratic equation using the quadratic formula [tex]\( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 1, b = -33, \)[/tex] and [tex]\( c = 38 \)[/tex]:

[tex]\[ r = \frac{33 \pm \sqrt{(-33)^2 - 4 \cdot 1 \cdot 38}}{2 \cdot 1} \][/tex]

[tex]\[ r = \frac{33 \pm \sqrt{1089 - 152}}{2} \][/tex]

[tex]\[ r = \frac{33 \pm \sqrt{937}}{2} \][/tex]

Thus, the solutions are:

[tex]\[ r = \frac{33 - \sqrt{937}}{2} \quad \text{or} \quad r = \frac{33 + \sqrt{937}}{2} \][/tex]

And considering the possibility of [tex]\( r = 0 \)[/tex]:

[tex]\[ r = 0, \quad \frac{33 - \sqrt{937}}{2}, \quad \frac{33 + \sqrt{937}}{2} \][/tex]

These are the solutions for the given equation.
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