2
Active contributors today

## A fan of country music plans to make a custom cd with 12 of her 28 favorite songs. How many different combinations of 12 songs are possible? Is it practical to make a different cd for each possible combination

Parzival S.
Featured 9 months ago

30,421,755 CDs. Realistic to make a CD of every combination? No. Practical? What is practicality to a true music lover?!?!?!

#### Explanation:

We have 28 songs and we're going to find how many ways we can put sets of 12 songs onto CDs. Since order doesn't matter (having songs 1 - 12 on a CD is the same as having songs 12 - 1 on the CD), we'll use the Combination calculation, which is:

C_(28, 12)=(28!)/((12!)(28-12)!)=(28!)/((12!)(16!)

And now let's evaluate this fraction:

(28xx27xx26xx25xx24xx23xx22xx21xx20xx19xx18xx17xxcancel(16!))/((12!)cancel(16!)

$\frac{28 \times 27 \times 26 \times 25 \times 24 \times 23 \times 22 \times 21 \times 20 \times 19 \times 18 \times 17}{12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2}$

There's a lot of cancelations, so I'll use colours to help keep track:

(cancel(color(red)28)xxcancel(color(blue)(27))xx26xxcancel(color(orange)(25))^5xxcancel(color(green)(24))xx23xxcancel(color(pink)(22))^2xx21xxcancel(color(brown)(20))^2xx19xxcancel(color(tan)(18))^3xx17)/(cancel(color(green)(12))xxcancel(color(pink)(11))xxcancel(color(brown)(10))xxcancel(color(blue)(9))xx8xxcancel(color(red)(7))xxcancel(color(tan)(6))xxcancel(color(orange)(5))xxcancel(color(red)4)xxcancel(color(blue)3)xxcancel(color(green)(2))

Let's clean this up and do one more round... I'm going to express 26 as $13 \times 2$

$\frac{13 \times \cancel{2} \times 5 \times 23 \times \cancel{2} \times 21 \times \cancel{2} \times 19 \times 3 \times 17}{\cancel{8}}$

$\left(13 \times 5 \times 23 \times 21 \times 19 \times 3 \times 17\right) = 30 , 421 , 755$

(To avoid the long and involved calculations, you can use the Combination calculator tool here: http://www.calculatorsoup.com/calculators/discretemathematics/combinations.php)

## How to prove this equation?

Shwetank Mauria
Featured 7 months ago

Use the identity $\left(\begin{matrix}n \\ k\end{matrix}\right) = \left(\begin{matrix}n - 1 \\ k\end{matrix}\right) + \left(\begin{matrix}n - 1 \\ k - 1\end{matrix}\right)$. See below for details.

#### Explanation:

Recall the identity $\left(\begin{matrix}n \\ k\end{matrix}\right) = \left(\begin{matrix}n - 1 \\ k\end{matrix}\right) + \left(\begin{matrix}n - 1 \\ k - 1\end{matrix}\right)$

To prove this as ((n),(k))=(n!)/(k!(n-k)!)

=(nxx(n-1)!)/(kxx(k-1)!(n-k)(n-k-1)!)

=((n-1)!)/((k-1)!(n-k-1)!)[n/(k(n-k))]

=((n-1)!)/((k-1)!(n-k-1)!)[((n-k)+k)/(k(n-k))]

=((n-1)!)/((k-1)!(n-k-1)!)[1/k+1/(n-k)]

=((n-1)!)/(k(k-1)!(n-k-1)!)+((n-1)!)/(k(k-1)!(n-k)(n-k-1)!)

=((n-1)!)/(k!(n-k-1)!)+((n-1)!)/(k!(n-k)!)=((n-1),(k))+((n-1),(k-1))

Hence $\left(\begin{matrix}n \\ k\end{matrix}\right) = \left(\begin{matrix}n - 1 \\ k\end{matrix}\right) + \left(\begin{matrix}n - 1 \\ k - 1\end{matrix}\right)$, bur from this we get

$\left(\begin{matrix}n - 1 \\ k - 1\end{matrix}\right) = \left(\begin{matrix}n - 2 \\ k - 1\end{matrix}\right) + \left(\begin{matrix}n - 2 \\ k - 2\end{matrix}\right)$

Hence $\left(\begin{matrix}n \\ k\end{matrix}\right) = \left(\begin{matrix}n - 1 \\ k\end{matrix}\right) + \left(\begin{matrix}n - 2 \\ k - 1\end{matrix}\right) + \left(\begin{matrix}n - 2 \\ k - 2\end{matrix}\right)$

Now similarly split $\left(\begin{matrix}n - 2 \\ k - 2\end{matrix}\right)$ into

$\left(\begin{matrix}n - 2 \\ k - 2\end{matrix}\right) = \left(\begin{matrix}n - 3 \\ k - 2\end{matrix}\right) + \left(\begin{matrix}n - 3 \\ k - 3\end{matrix}\right)$ and

$\left(\begin{matrix}n \\ k\end{matrix}\right) = \left(\begin{matrix}n - 1 \\ k\end{matrix}\right) + \left(\begin{matrix}n - 2 \\ k - 1\end{matrix}\right) + \left(\begin{matrix}n - 3 \\ k - 2\end{matrix}\right) + \left(\begin{matrix}n - 3 \\ k - 3\end{matrix}\right)$

Going on similarly, we get

$\left(\begin{matrix}n \\ k\end{matrix}\right) = {\sum}_{l = k}^{n} \left(\begin{matrix}l - 1 \\ k - 1\end{matrix}\right)$

## What is the probability that the deal of a five-card hand provides exactly one ace?

Parzival S.
Featured 6 months ago

P=(778,320)/(2,598,960)~=29.9%

#### Explanation:

An alternative way to do this is to use equations for combinations, the general formula for which is:

C_(n,k)=(n!)/((k)!(n-k)!) with $n = \text{population", k="picks}$

We want our hand to have exactly one Ace. Since there are four aces in a deck, we can set $n = 4 , k = 1$, and so:

${C}_{4 , 1}$

But we also need to account for the other four cards in our hand. There are 48 cards to pick from, so we can set $n = 48 , k = 4$, and so:

${C}_{48 , 4}$

We multiply them together to find the total number of ways we can get exactly one Ace in our hand:

C_(4,1)xxC_(48,4)=(4!)/((1)!(4-1)!)(48!)/((4)!(48-4)!)=>

(cancel(4!)(48!))/((3!)cancel(4!)(44!))=(cancel(48)^8xx47xx46xx45xxcancel44!)/(cancel(3xx2)xxcancel44!)=>

$8 \times 47 \times 46 \times 45 = 778 , 320$

~~~~~

To figure out the probability, we also need to know the total number of 5-card hands possible:

C_(52,5)=(52!)/((5)!(52-5)!)=(52!)/((5!)(47!))

Let's evaluate it!

(52xx51xxcancelcolor(orange)(50)^10xx49xxcancelcolor(red)48^2xxcancelcolor(brown)(47!))/(cancelcolor(orange)5xxcancelcolor(red)(4xx3xx2)xxcancelcolor(brown)(47!))=52xx51xx10xx49xx2=2,598,960

And so:

P=(778,320)/(2,598,960)~=29.9%

## What is 'z' for an 88% confidence interval?

Geoff K.
Featured 3 months ago

From software: $z = 1.554774 .$
From table lookup: $z \approx 1.56 .$

#### Explanation:

If we seek an 88% confidence interval, that means we only want a 12% chance that our interval does not contain the true value. Assuming a two-sided test, that means we want a 6% chance attributed to each tail of the $Z$-distribution. Thus, we seek the ${z}_{\alpha / 2}$ value of ${z}_{0.06}$.

This $z$ value at $\alpha / 2 = 0.06$ is the coordinate of the $Z$-curve that has 6% of the distribution's area to its right, and thus 94% of the area to its left. We find this $z$-value by reverse-lookup in a $z$-table.

Find the closest value in the table to 0.9400 as you can, then see what its row and column is. From observation, we see that 0.9394 and 0.9406 are in the table with $z$-values of 1.55 and 1.56 respectively, and so to err on the side of caution, we'll choose the value that gives us a wider interval, $z = 1.56 .$

Note: We could also get an answer from software like R, by typing the command $\text{qt(0.94, Inf)}$, which would give us a more precise value of 1.554774.

## A class contains 5 boys and 6 girls. The teacher is told that 3 of the students can go on a school trip, and selects them at random. What is the probability that 3 members of the same gender are selected?

Parzival S.
Featured 2 weeks ago

30/165=6/33=0.bar(18)~=0.18=18%

#### Explanation:

Let's find the number of ways all boys will go, then find the number of ways all girls will go, then add them together. Then we'll find the probability.

We're working with combinations (we don't care in what order the kids are picked). That general formula is:

C_(n,k)=(n!)/((k!)(n-k)!) with $n = \text{population", k="picks}$

I'll use $\textcolor{b l u e}{\text{blue for boys}}$ and $\textcolor{red}{\text{red for girls}}$

All boys

From the population of 5 boys, 3 are picked. From the population of girls, none are picked:

color(blue)(C_(5,3))xxcolor(red)(C_(6,0))=color(blue)((5!)/((3!)(5-3)!))xxcolor(red)((6!)/((0!)(6-0)!))=color(blue)((5!)/(3!2!))xxcolor(red)((6!)/(0!6!))=color(blue)(120/12)xxcolor(red)(1)=10

All girls

From the population of 6 girls, 3 are picked. From the population of boys, none are picked:

color(blue)(C_(5,0))xxcolor(red)(C_(6,3))=color(blue)((5!)/((0!)(5-0)!))xxcolor(red)((6!)/((3!)(6-3)!))=color(blue)((5!)/(0!5!))xxcolor(red)((6!)/(3!3!))=color(blue)(1)xxcolor(red)(720/36)=20

Total ways

$10 + 20 = 30$

Probability

We now have the total number of ways to meet the condition of having 3 of the same sex go on the field trip. How many different ways can we select 11 children to go? There are 11 children in total and we're picking 3:

C_(11,3)=(11!)/((3!)(11-3)!)=(11!)/(3!8!)=(11xx10xx9xx8!)/(6xx8!)=990/6=165

And so that gives us:

30/165=6/33=0.bar(18)~=0.18=18%

## ⁿ⁺¹c₅+ⁿ⁺²c₅+ⁿ⁺³c₅+ⁿ⁺⁴c₅=?

Parzival S.
Featured 2 weeks ago

$\implies \left(\frac{1}{60}\right) \left(2 {n}^{5} + 5 {n}^{4} + 20 {n}^{3} + 25 {n}^{2} + 8 n\right)$

#### Explanation:

We're adding combinations, where the general formula is:

C_(n,k)=(n!)/((k!)(n-k)!) with $n = \text{population", k="picks}$

Let's substitute in values:

ⁿ⁺¹c₅+ⁿ⁺²c₅+ⁿ⁺³c₅+ⁿ⁺⁴c₅

=>((n+1)!)/((5!)((n+1)-5)!)+((n+2)!)/((5!)((n+2)-5)!)+((n+3)!)/((5!)((n+3)-5)!)+((n+4)!)/((5!)((n+4)-5)!)

=>((n+1)!)/((5!)(n-4)!)+((n+2)!)/((5!)(n-3)!)+((n+3)!)/((5!)(n-2)!)+((n+4)!)/((5!)(n-1)!)

I'm going to simplify each fraction first before trying to combine them. Remember that n! = n xx (n-1) xx (n-2) xx ... xx 1 and so:

=>((n+1)(n)(n-1)(n-2)(n-3)(n-4)!)/((5!)(n-4)!)+((n+2)(n+1)(n)(n-1)(n-2)(n-3)!)/((5!)(n-3)!)+((n+3)(n+2)(n+1)(n)(n-1)(n-2)!)/((5!)(n-2)!)+((n+4)(n+3)(n+2)(n+1)(n)(n-1)!)/((5!)(n-1)!)

=>((n+1)(n)(n-1)(n-2)(n-3)(n-4)!)/((5!)(n-4)!)+((n+2)(n+1)(n)(n-1)(n-2)(n-3)!)/((5!)(n-3)!)+((n+3)(n+2)(n+1)(n)(n-1)(n-2)!)/((5!)(n-2)!)+((n+4)(n+3)(n+2)(n+1)(n)(n-1)!)/((5!)(n-1)!)

Know that 5! = 120, and so:

$\implies \frac{1}{120} \left(\left(n + 1\right) \left(n\right) \left(n - 1\right) \left(n - 2\right) \left(n - 3\right) + \left(n + 2\right) \left(n + 1\right) \left(n\right) \left(n - 1\right) \left(n - 2\right) + \left(n + 3\right) \left(n + 2\right) \left(n + 1\right) \left(n\right) \left(n - 1\right) + \left(n + 4\right) \left(n + 3\right) \left(n + 2\right) \left(n + 1\right) \left(n\right)\right)$

And we can simplify a bit more:

$\implies \left(\frac{\left(n + 1\right) \left(n\right)}{120}\right) \left(\left(n - 1\right) \left(n - 2\right) \left(n - 3\right) + \left(n + 2\right) \left(n - 1\right) \left(n - 2\right) + \left(n + 3\right) \left(n + 2\right) \left(n - 1\right) + \left(n + 4\right) \left(n + 3\right) \left(n + 2\right)\right)$

$\implies \left(\frac{\left(n + 1\right) \left(n\right)}{120}\right) \left(\left(n - 1\right) \left(n - 2\right) \left(n - 3\right) + \left(n + 2\right) \left(n - 1\right) \left(n - 2\right) + \left(n + 3\right) \left(n + 2\right) \left(n - 1\right) + \left(n + 4\right) \left(n + 3\right) \left(n + 2\right)\right)$

=>((n^2+n)/120)((n^3-6n^2+11n-6)+(n^3-n^2-4n+4)+(n^3+4n^2+n-6)+(n^3+9n^2+26n+24)

$\implies \left(\frac{{n}^{2} + n}{120}\right) \left(4 {n}^{3} + 6 {n}^{2} + 34 n + 16\right)$

$\implies \left(\frac{1}{60}\right) \left({n}^{2} + n\right) \left(2 {n}^{3} + 3 {n}^{2} + 17 n + 8\right)$

$\implies \left(\frac{1}{60}\right) \left(2 {n}^{5} + 5 {n}^{4} + 20 {n}^{3} + 25 {n}^{2} + 8 n\right)$

And now to see if I got the math right. Suppose $n = 5$. Taking the original expression, we'd calculate:

${C}_{6 , 5} + {C}_{7 , 5} + {C}_{8 , 5} + {C}_{9 , 5} = 6 + 21 + 56 + 126 = 209$

And from the expression we've found:

$\left(\frac{1}{60}\right) \left(2 {\left(5\right)}^{5} + 5 {\left(5\right)}^{4} + 20 {\left(5\right)}^{3} + 25 {\left(5\right)}^{2} + 8 \left(5\right)\right)$

$\left(\frac{1}{60}\right) \left(6250 + 3125 + 2500 + 625 + 40\right)$

(1/60)(12540)=209color(white)(000)color(green)root

##### Questions
• · Yesterday
• · 2 days ago
• · 4 days ago
• · 4 days ago
• · 4 days ago
• · 5 days ago
• · 1 week ago · in Measures of Center
• · 1 week ago · in What is Statistics?
• · 1 week ago
• · 1 week ago
• 1 week ago · in Boxplots
• · 1 week ago