Question

10c6+10c7+11c8 simplify

Answer 1

Given 10c6 + 10c7 + 11c8 ------------ (1)

We know that ncr = n!/r!(n-r)!

10c6 = 10!/6!(10! - 6!)

        = 10!/6! * 4!

        = 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1/ 6 * 5 * 4 * 3 * 2 * 1 * 4 * 3 * 2 * 1

        = 10 * 9 * 8 * 7/4 * 3 * 2 * 1 

        = 210.   ------ (2)



10c7 = 10!/7!(10 - 7!)

        = 10!/7! * 3!

       = 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1/7 * 6 * 5 * 4 * 3 * 2 * 1 * 3 * 2 * 1

       = 10 * 9 * 8/3 * 2 * 1

       = 120.  ---------- (3)


11c8 = 11!/8!(11 - 8)!

        = 11!/8! * 3!

        = 11 * 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1/ 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 * 3 * 2 * 1

       = 11 * 10 * 9/3 * 2 * 1

       = 165   --------- (4).



Substitute (2),(3),(4) in (1), we get


10c6 + 10c7 + 11c8 = 210 + 120 + 165

                               = 495.



Hope this helps!

Answer 2

my dear friend
in binomial their is a property which states that ncr+ nc(r-1) = (n+1)cr

so applying this we get
10c6+10c7= 11c7
which in turn adds with 11c8
thus 11c7+11c8 = 12c8
which is (12*11*10*9)/(4*3*2*1)
= 495