Question 7
(a) A 5-character password is to be formed from the following 13 characters.
| Letters | A | B | C | D | E |
| Numbers | 9 | 8 | 7 | 6 | 5 |
| Symbols | $*$ | $\#$ | $!$ |
No character may be used more than once in any password.
(i) Find the number of possible passwords that can be formed. [1]
(ii) Find the number of possible passwords that contain at least one symbol. [2]
(b) Given that $ 16 \times { }^n \mathrm{C}_{12} = (n-10)\times { }^{n+1}\mathrm{C}_{11} $, find the value of $n$. [3]
Solution
Step (a) - Part (i): Number of possible passwords
We are tasked with forming a 5-character password using 13 available characters, which consist of letters, numbers, and symbols.
These characters are:
$$ \text{Letters: } A,B,C,D,E \quad \text{(5 letters)} $$ $$ \text{Numbers: } 9,8,7,6,5 \quad \text{(5 numbers)} $$ $$ \text{Symbols: } *,\#,! \quad \text{(3 symbols)} $$The total number of characters available is $13$. Since no character may be used more than once in any password, we use permutations.
The number of possible passwords is:
$$ P(13,5) = \frac{13!}{(13-5)!} = \frac{13!}{8!} $$Compute:
$$ P(13,5) = 13\times12\times11\times10\times9 = 154440 $$Thus, the number of possible passwords is $ \boxed{154440} $.
Step (a) - Part (ii): Number of passwords with at least one symbol
Use complementary counting.
Total passwords:
$$ 154440 $$Now find the number of passwords with no symbols. Only the 10 letters and numbers are used.
$$ P(10,5) = \frac{10!}{(10-5)!} = \frac{10!}{5!} $$Compute:
$$ P(10,5) = 10\times9\times8\times7\times6 = 30240 $$Therefore:
$$ 154440-30240=124200 $$Thus, the number of passwords containing at least one symbol is $ \boxed{124200} $.
Step (b): Solve for $n$
We are given:
$$ 16\times {^n\mathrm{C}_{12}} = (n-10)\times {^{n+1}\mathrm{C}_{11}} $$Recall:
$$ {^n\mathrm{C}_k} = \frac{n!}{k!(n-k)!} $$Substitute into the equation:
$$ 16\times\frac{n!}{12!(n-12)!} = (n-10)\times \frac{(n+1)!}{11!(n-10)!} $$Since $ (n+1)!=(n+1)n! $:
$$ 16\times\frac{n!}{12!(n-12)!} = (n-10)\times \frac{(n+1)(n!)}{11!(n-10)!} $$Cancel $n!$:
$$ 16\times\frac{1}{12!(n-12)!} = (n-10)\times \frac{(n+1)}{11!(n-10)!} $$Simplify:
$$ 16\times\frac{1}{12!(n-12)!} = \frac{(n-10)(n+1)}{11!(n-10)!} $$Multiply both sides by $ 12!(n-12)! $:
$$ 16 = \frac{(n-10)(n+1)\times12!(n-12)!} {11!(n-10)!} $$ $$ 16 = \frac{ (n-10)(n+1)\times12\times11!(n-12)! }{ 11!(n-10)(n-11)(n-12)! } $$ $$ 16 = \frac{12(n+1)}{n-11} $$Cross multiply:
$$ 16(n-11)=12(n+1) $$ $$ 16n-176=12n+12 $$ $$ 4n=188 $$ $$ \boxed{n=47} $$Question 8
The diagram shows part of the curve $ y=2-\frac{3}{x-1} $ and the straight line $ 6y=9-2x. $ The curve intersects the $x$-axis at point $A$ and the line at point $B$. The line intersects the $x$-axis at point $C$. Find the area of the shaded region $ABC$, giving your answer in the form $ p+\ln q, $ where $p$ and $q$ are rational numbers. [11]
Solution
Step 1: Find the equations of the curve and line
The equation of the curve is:
$$ y=2-\frac{3}{x-1} $$The equation of the line is:
$$ 6y=9-2x $$Solving for $y$:
$$ y=\frac{9-2x}{6} $$Step 2: Find point $A$ where the curve meets the $x$-axis
At the $x$-axis, $ y=0. $
$$ 0=2-\frac{3}{x-1} $$Solve for $x$:
$$ \frac{3}{x-1}=2 $$ $$ x-1=\frac{3}{2} $$ $$ x=\frac{5}{2} $$Therefore,
$$ A\left(\frac{5}{2},0\right) $$Step 3: Find point $B$ where the curve and line intersect
Set the equations equal:
$$ 2-\frac{3}{x-1}=\frac{9-2x}{6} $$Multiply both sides by $6$:
$$ 6\left(2-\frac{3}{x-1}\right)=9-2x $$ $$ 12-\frac{18}{x-1}=9-2x $$Rearrange:
$$ 3=-2x+\frac{18}{x-1} $$Multiply by $(x-1)$:
$$ 3(x-1)=-2x(x-1)+18 $$Expand:
$$ 3x-3=-2x^2+2x+18 $$ $$ 2x^2+x-21=0 $$Use the quadratic formula:
$$ x=\frac{-1\pm\sqrt{1^2-4(2)(-21)}}{2(2)} $$ $$ x=\frac{-1\pm\sqrt{169}}{4} $$ $$ x=\frac{-1\pm13}{4} $$Hence:
$$ x=3 \quad \text{or} \quad x=-3.5 $$Take the positive value:
$$ x=3 $$Substitute into the line equation:
$$ y=\frac{9-2(3)}{6} =\frac{3}{6} =\frac{1}{2} $$Therefore,
$$ B\left(3,\frac{1}{2}\right) $$Step 4: Find point $C$ where the line meets the $x$-axis
At the $x$-axis, $ y=0. $
$$ 0=\frac{9-2x}{6} $$ $$ 9-2x=0 $$ $$ x=\frac{9}{2} $$Therefore,
$$ C\left(\frac{9}{2},0\right) $$Step 5: Find the area of the shaded region
The required area is:
$$ \text{Area} = \int_{2.5}^{3} \left( 2-\frac{3}{x-1} \right)dx + \int_{3}^{4.5} \frac{9-2x}{6}\,dx $$First Integral
$$ \int_{2.5}^{3} \left( 2-\frac{3}{x-1} \right)dx $$ $$ = \int_{2.5}^{3}2\,dx - \int_{2.5}^{3}\frac{3}{x-1}\,dx $$ $$ = \left[2x\right]_{2.5}^{3} - 3\ln|x-1|\Big|_{2.5}^{3} $$ $$ = \left[2(3)-2(2.5)\right] - 3\ln\left(\frac{3-1}{2.5-1}\right) $$ $$ = 1-3\ln\left(\frac{4}{3}\right) $$Second Integral
$$ \int_{3}^{4.5}\frac{9-2x}{6}\,dx $$ $$ = \frac{1}{6} \int_{3}^{4.5}(9-2x)\,dx $$ $$ = \frac{1}{6} \left[9x-x^2\right]_{3}^{4.5} $$ $$ = \frac{1}{6} \left[ (9\times4.5-(4.5)^2) - (9\times3-3^2) \right] $$ $$ = \frac{1}{6} \left[ (40.5-20.25) - (27-9) \right] $$ $$ = \frac{1}{6}(20.25-18) $$ $$ = \frac{1}{6}(2.25) =0.375 =\frac{3}{8} $$Total Area
$$ \text{Area} = \left( 1-3\ln\left(\frac{4}{3}\right) \right) + \frac{3}{8} $$ $$ = \frac{11}{8} - 3\ln\left(\frac{4}{3}\right) $$ $$ = \frac{11}{8} + 3\ln\left(\frac{3}{4}\right) $$ $$ = \frac{11}{8} + \ln\left(\frac{3}{4}\right)^3 $$ $$ = \frac{11}{8} + \ln\left(\frac{27}{64}\right) $$Thus,
$$ \boxed{ \frac{11}{8} + \ln\frac{27}{64} } $$
Post a Comment