Given that ^nC2 = 66, find the value of n.

Question

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here are the solutions to the remaining problems.

1.
Given that $^nC_2 = 66$ , find the value of $n$ .

Write down the formula for combinations.
$^nC_r = \frac{n!}{r!(n-r)!}$
Substitute $r=2$ into the formula.
$^nC_2 = \frac{n!}{2!(n-2)!}$
Expand $n!$ as $n(n-1)(n-2)!$ and simplify.
$^nC_2 = \frac{n(n-1)(n-2)!}{2 \times 1 \times (n-2)!} = \frac{n(n-1)}{2}$
Set the expression equal to 66 and solve for $n$ .
$\frac{n(n-1)}{2} = 66$ $n(n-1) = 132$ $n^2 - n - 132 = 0$
Factor the quadratic equation.
$(n-12)(n+11) = 0$
This gives two possible values for $n$ :
$n-12 = 0 \implies n = 12$ $n+11 = 0 \implies n = -11$
Since $n$ must be a positive integer for combinations, we discard $n=-11$ .
The value of $n$ is $\boxed{12}$ .

2.
Using the trapezium rule with five ordinates, evaluate $\int_0^2 \frac{4}{1+x^2} dx$ , correct to two decimal places.
Determine the number of strips ( $n$ ) and the width of each strip ( $h$ ).
Number of ordinates = 5, so the number of strips $n = 5-1 = 4$ .
The interval is from $a=0$ to $b=2$ .
The strip width $h = \frac{b-a}{n} = \frac{2-0}{4} = \frac{2}{4} = 0.5$ .
Calculate the $x$ -values for each ordinate and their corresponding $y$ -values, $y = f(x) = \frac{4}{1+x^2}$ .
- $x_0 = 0 \implies y_0 = \frac{4}{1+0^2} = \frac{4}{1} = 4$
- $x_1 = 0.5 \implies y_1 = \frac{4}{1+(0.5)^2} = \frac{4}{1+0.25} = \frac{4}{1.25} = 3.2$
- $x_2 = 1.0 \implies y_2 = \frac{4}{1+1.0^2} = \frac{4}{1+1} = \frac{4}{2} = 2$
- $x_3 = 1.5 \implies y_3 = \frac{4}{1+1.5^2} = \frac{4}{1+2.25} = \frac{4}{3.25} \approx 1.230769$
- $x_4 = 2.0 \implies y_4 = \frac{4}{1+2.0^2} = \frac{4}{1+4} = \frac{4}{5} = 0.8$
Apply the trapezium rule formula
$\int_a^b y \, dx \approx \frac{h}{2} [ (y_0 + y_n) + 2(y_1 + y_2 + \dots + y_{n-1}) ]$
Substitute the values:
$\int_0^2 \frac{4}{1+x^2} dx \approx \frac{0.5}{2} [ (y_0 + y_4) + 2(y_1 + y_2 + y_3) ]$ $\approx 0.25 [ (4 + 0.8) + 2(3.2 + 2 + 1.230769) ]$ $\approx 0.25 [ 4.8 + 2(6.430769) ]$ $\approx 0.25 [ 4.8 + 12.861538 ]$ $\approx 0.25 [ 17.661538 ]$ $\approx 4.4153845$
Round the result to two decimal places.
$\approx 4.42$
The value of the integral is approximately $\boxed{4.42}$ .

Given that ^nC2 = 66, find the value of n.

Answer

Step-by-step solution

Write down the formula for combinations.

Substitute $r=2$ into the formula.

Expand $n!$ as $n(n-1)(n-2)!$ and simplify.

Set the expression equal to 66 and solve for $n$ .

Factor the quadratic equation.

Determine the number of strips ( $n$ ) and the width of each strip ( $h$ ).

Calculate the $x$ -values for each ordinate and their corresponding $y$ -values, $y = f(x) = \frac{4}{1+x^2}$ .

Apply the trapezium rule formula

Round the result to two decimal places.

Try one yourself

Still stuck on this one?

More Mathematics Questions

Given that ^nC2 = 66, find the value of n.

Answer

Step-by-step solution

Write down the formula for combinations.

Substitute r=2r=2r=2 into the formula.

Expand n!n!n! as n(n−1)(n−2)!n(n-1)(n-2)!n(n−1)(n−2)! and simplify.

Set the expression equal to 66 and solve for nnn.

Factor the quadratic equation.

Determine the number of strips (nnn) and the width of each strip (hhh).

Calculate the xxx-values for each ordinate and their corresponding yyy-values, y=f(x)=41+x2y = f(x) = \frac{4}{1+x^2}y=f(x)=1+x24​.

Apply the trapezium rule formula

Round the result to two decimal places.

Try one yourself

Still stuck on this one?

More Mathematics Questions

Substitute $r=2$ into the formula.

Expand $n!$ as $n(n-1)(n-2)!$ and simplify.

Set the expression equal to 66 and solve for $n$ .

Determine the number of strips ( $n$ ) and the width of each strip ( $h$ ).

Calculate the $x$ -values for each ordinate and their corresponding $y$ -values, $y = f(x) = \frac{4}{1+x^2}$ .