Please help! Permutation and Combination.A committee of 5 is to be formed out of 6 men & 4 women.In how many ways can this be done when:?

The total number of committees of 5 is C(10,5). This is equal to the sum of the committees formed with 0, 1, 2, 3, 4 women and 5, 4, 3, 2, 1 men:

total committees of 5 = (committees with 0 W, 5 M) + (committees with 1 W, 4M) + (comitted with 2 W, 3 M) + (committees with 3 W, 2 M) + (committees with 4 W, 1 M)

or equivilantly:

C(10,5) = C(4,0)C(6,5) + C(4,1)C(6,4) + C(4,2)C(6,3) + C(4,3)C(6,2) + C(4,4)C(6,1)

[this is called the Chu-Vandermonde Identity and if we evaluate the binomial coefficients we get the equation below]

252 = 6 + 60 + 120 + 60 + 6

i) This is the sum of the first 3 terms = 6+60+120 = 186 <----
ii) This is the sum of the last 3 terms = 120+60+6 = 186 <----

(i)
at least 2 women
W M
2 3 ---> (4C2)(6C3) = (6) (20) = 120
3 2 ---> (4C3)(6C1) = (4)(6) = 24
4 1 ----> (4C4) (6C0) = 1

number of ways with at least 2 women = 120+24+1 = 145

(ii)
at most 2 women
W M
0 5 ----> (4C0) (6C5) = (1) (6) = 6
1 4 ----> (4C1)(6C4) = (4) (15) = 60
2 3 ----> (4C2) (6C3) = (6) (20) = 120

number of ways with at most 2 women = 6+60+120 = 186