Of 19 possible books, you plan to take 6 with you on vacation. How many different collections of 6 books can you take?

Mathematics

1 answer:

mash [69]2 years ago

7 0

Answer:

27132

Step-by-step explanation:

Assuming the order does not matter, since it will always be a mixture of the same 6 books, we can calculate a combination for these outcomes.

The combination formula is:

nCr = n! ÷ (n-r)!r!

, where ! represents factorials, n represents the total number of items, and r represents the number of items being chosen.

So 19! ÷ (19-6)!6! = 19! ÷ 13! • 6!

19 • 18 • 17 • 16 • 15 • 14 ÷ 6 • 5 • 4 • 3 • 2

= 27132 different combinations of the 6 books.

If you use a calculator, find nCr, plug in 19 for n, and 6 for r.

You might be interested in

V<img src="https://tex.z-dn.net/?f=%5Cfrac%7B3%7D%7B4%7D%20%2A%202%5Cfrac%7B1%7D%7B3%7D" id="TexFormula1" title="\frac{3}{4} * 2

myrzilka [38]

Answer:

$1\frac{3}{4}$

Step-by-step explanation:

In order to find the answer, you first need to convert the mixed number (2 1/3) into a improper fraction and then multiply.

$\frac{3}{4} \times2\frac{1}{3}$

$2\frac{1}{3} =3\times2=6+1=7=\frac{7}{3}$

$\frac{3}{4} \times\frac{7}{3}$

$3\times7=21$

$4\times3=12$

$=\frac{21}{12} =\frac{7}{4} =1\frac{3}{4}$

$=1\frac{3}{4}$

Hope this helps.

7 0

2 years ago

Answer if you know

Arlecino [84]

I've attached a plot of one such cross-section (orange) over the region in the x-y plane (blue), including the bounding curves (red). (I've set $y=1.25$ for this example.)

The length of each cross section (the side lying in the base) has length determined by the horizontal distance $x$ between the y-axis $x=0$ and the curve $y=\sqrt x$ . In terms of $y$ , this distance is $x=y^2$ . The height of each cross section is twice the value of $y$ , so the area of each rectangular cross section should be $2y^3$ .

This means the volume would be given by the integral

$\displaystyle\int_0^22y^3\,\mathrm dy$