how to solve combined factorial
example;
10C8
10/(10-8)!8! = 10.9.8.7.6.5.4.3.2.1/2!8! cancel 8 to 1 = pag x mo Yung naiwan na number which is 10.9 = 10x9/2x1 = 90/2 pag dividen mo = 45 that is the answer
example;
10C8
10/(10-8)!8! = 10.9.8.7.6.5.4.3.2.1/2!8! cancel 8 to 1 = pag x mo Yung naiwan na number which is 10.9 = 10x9/2x1 = 90/2 pag dividen mo = 45 that is the answer
Answers & Comments
Another way to solve "combination" problems involving factorials is to use the formula:
n C r = n!/[(n-r)! r!]
Where n is the total number of items, and r is the number of items being selected or arranged.
So, for example, to find 10C8, we plug in n=10 and r=8:
10C8 = 10!/[(10-8)!8!]
= 10!/[2!8!]
= (10x9x8x7x6x5x4x3x2x1)/[2x1x8x7x6x5x4x3x2x1]
= 90/2
= 45
Therefore, 10C8 is equal to 45.
Yes, you are correct! To expand a bit more on your explanation, we can solve the combination formula using factorials as follows:
nCr = n! / r!(n-r)!
where n is the total number of objects and r is the number of objects chosen.
In the example you provided, we have:
10C8 = 10! / 8!(10-8)!
= 10! / 8!2!
= 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 / (2 x 1)(8 x 7 x 6 x 5 x 4 x 3 x 2 x 1)
= (10 x 9) / 2
= 45
Therefore, 10C8 is equal to 45.