just write it out x+(x+4)+(x+8 )...+(x+36)=80
solve 10x+180=80, x=-10
for 40 times, 40(-10)+720=320.

or knowing that sum of 10 items is 80, multiply by 4 to solve for 40th term, 80*4=320

I think....

the answer to the second question is (4C2)/(10C3) = 1/6

evansbd wrote:I think....

The answer to the second question is (4C2)/(10C3) = 1/6

Well the question says at least 2 girls, means more than 2 will also be taken into account.

4C2 * 6C1 / 10C3

+

4C3*6C0 / 10C3

= 1/3


Let me if you think otherwise.

Definitely...I never accounted for all 3 girls being chosen

seweryn wrote:just write it out x+(x+4)+(x+8 )...+(x+36)=80
solve 10x+180=80, x=-10
for 40 times, 40(-10)+720=320.

or knowing that sum of 10 items is 80, multiply by 4 to solve for 40th term, 80*4=320

Sum of n terms = n/2[2a+(n-1)d]

x+(x+4)+(x+8 )...+(x+36)=80

Sum of first 10 terms

10/2[2x+(10-1)4] = 80

x = -10

Sum of first 40 terms

40/2[2(-10) + (40-1)4] => 2720

Let me know what you think.

Question 1:
Agreed with parallel_ chase

Sum of the first ten (i.e. x, x+ 4,X+8, ... X+36) or ( X+1x4,X+ 2x4,3x4....X+9x4)

Knowing that the sum of the first 10 numbers is 80
x= -10

Now for the first 40 numbers: In order to make it easier, let us isolate the x's and the multiples of 4

Hence,

eq. 1:40 x= -400

eq.2 (1X4+ 2X4+...... +39X4) equals to [(39X4)X40]/2 = 3120

3120-400= 2720

Of course you don`t have to go through all this presentation but there is a formula to be used in any sequence:

this formula specifies that for a number of n elements in a sequence, the sum of the elements is [n(n+1)]/2

To illustrate:

{1+2+3+4}= 10

Using the formula, the sum would be : [4(5)]/2= 10

Q.2

Since we are asked for AT LEAST two girls with blue eyes, the third spot can be taken by any of the rest of the girls.

Therefore, we add the combinations above mentioned to get the result of 1/3.