evaluate
8C5
Sorry, I do not understand your notation.
Do you perhaps mean the binomialcoefficient C(8,5) ?
That would be
8!
-------
(5!)(3!)
which is
8*7*6
-------
3*2
which is
8*7
which is
56
Damon, back in the olden days, in the days of antiquity when I was teaching that stuff we often used the notation
8C5
and it meant C(8,5)
Some older text book would have used the notation
8P5 to stand for P(8,5)
To evaluate 8C5, which represents the number of combinations of 8 items taken 5 at a time, you can use the formula for combinations:
nCr = n! / (r!(n-r)!)
In this case, n = 8 and r = 5.
Plugging these values into the formula, you get:
8C5 = 8! / (5!(8-5)!)
= 8! / (5!3!)
Using factorials:
8! = 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1
5! = 5 x 4 x 3 x 2 x 1
3! = 3 x 2 x 1
Plugging in the values, you have:
8C5 = (8 x 7 x 6 x 5 x 4 x 3 x 2 x 1) / [(5 x 4 x 3 x 2 x 1) x (3 x 2 x 1)]
= (8 x 7 x 6) / (3 x 2 x 1)
Simplifying:
8C5 = (336) / (6)
= 56
Therefore, 8C5 is equal to 56.
To evaluate "8C5", we need to use the binomial coefficient formula:
nCr = n! / (r! * (n - r)!)
In this case, n = 8 and r = 5.
Now let's calculate it step-by-step:
1. Compute 8 factorial (8!):
8! = 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 = 40,320
2. Calculate 5 factorial (5!):
5! = 5 * 4 * 3 * 2 * 1 = 120
3. Calculate the factorial of (8-5) = 3:
3! = 3 * 2 * 1 = 6
4. Substitute the values into the formula:
8C5 = 8! / (5! * (8-5)!)
= 40,320 / (120 * 6)
= 56
Therefore, 8C5 is equal to 56.