Answered

Welcome to Westonci.ca, where finding answers to your questions is made simple by our community of experts. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.

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.
Thanks for stopping by. We are committed to providing the best answers for all your questions. See you again soon. Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. Westonci.ca is your go-to source for reliable answers. Return soon for more expert insights.