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2 years ago

How many ways to sit 10 people around a table?

2 answers:

8 0

362,880. You can solve this by doing 9! (9*8*7*6*5*4*3*2*1). You don't do it by ten because one of the spots has to be fixated.

murzikaleks [220]2 years ago

7 0

10 factorial times to sit 10 people around table.
Factorial is defined as n! = n*(n-1)*(n-2)*...*2*1

10*9*8*7*6*5*4*3*2*1=3628800 ways to sit people around table.

You might be interested in

Expansion Numerically Impractical. Show that the computation of an nth-order determinant by expansion involves multiplications,

posledela

Answer:

number of multiplies is n!
n=10, 3.6 ms
n=15, 21.8 min
n=20, 77.09 yr
n=25, 4.9×10^8 yr

Step-by-step explanation:

Expansion of a 2×2 determinant requires 2 multiplications. Expansion of an n×n determinant multiplies each of the n elements of a row or column by its (n-1)×(n-1) cofactor determinant. Then the number of multiplies is ...

mpy[n] = n·mp[n-1]

mpy[2] = 2

So, ...

mpy[n] = n! . . . n ≥ 2

If each multiplication takes 1 nanosecond, then a 10×10 matrix requires ...

10! × 10^-9 s ≈ 0.0036288 s ≈ 0.004 s . . . for 10×10

Then the larger matrices take ...

n=15, 15! × 10^-9 ≈ 1307.67 s ≈ 21.8 min

n=20, 20! × 10^-9 ≈ 2.4329×10^9 s ≈ 77.09 years

n=25, 25! × 10^-9 ≈ 1.55112×10^16 s ≈ 4.915×10^8 years

_____

For the shorter time periods (less than 100 years), we use 365.25 days per year.

For the longer time periods (more than 400 years), we use 365.2425 days per year.

8 0

3 years ago

What are the zeros of the quadratic function f(x) = 2x2 – 10x – 3?

Ipatiy [6.2K]

Answer:

$x = \frac{ 5 \ + \ \sqrt{31}}{2} \ , \ x = \frac{ 5 \ - \ \sqrt{31}}{2}$

Step-by-step explanation:

$2x^2 - 10x - 3 = 0 \\\\a = 2 \ , b = - 10 \ , \ c = - 3 \\\\x = \frac{-b^2\ \pm \ \sqrt{b^2 - 4ac}}{2a}\\\\x = \frac{10 \ \pm \sqrt{(-10)^2 - ( 4 \times 2 \times -3)} }{2 \times 2}\\\\x = \frac{10 \ \pm \sqrt{(100 - ( -24 )} }{4}\\\\x = \frac{10 \ \pm \sqrt{(100 + 24 } }{4}\\\\x = \frac{ 10 \ \pm \sqrt{124}}{4}\\\\x = \frac{ 10 \ \pm \sqrt{4 \times 31}}{4}\\\\x = \frac{ 10 \ \pm \sqrt{2^2 \times 31}}{4}\\\\x = \frac{ 10 \ \pm2 \sqrt{31}}{4}\\\\x = \frac{ 5 \ \pm\sqrt{31}}{2}\\\\$

$x = \frac{ 5 \ + \ \sqrt{31}}{2} \ , \ x = \frac{ 5 \ - \ \sqrt{31}}{2}$

3 0

2 years ago

How to write a slope equation using y intercept as 6 and the slope as -2

Bingel [31]

The slope equation is y=mx+b where m is the slope and b is the y intercept. After plugging the numbers into the equation you should get y=-2x+6. Hope this helps!

4 0

2 years ago

ΔEFG is located at E (0, 0), F (−7, 4), and G (0, 8). Which statement correctly classifies ΔEFG?

MariettaO [177]

Answer:

ΔEFG is an isosceles triangle.

Step-by-step explanation:

Given:

E (0, 0),

F (−7, 4),

G (0, 8)

ΔEFG