A computer programming team has 13 members. a. How many ways can a group of seven be chosen to work on a project? b. Suppose seven team members are women and six are men. (i) How many groups of seven can be chosen that contain four women and three men? (ii) How many groups of seven can be chosen that contain at least one man? (iii) How many groups of seven can be chosen that contain at most three women? c. Suppose two team members refuse to work together on projects. How many groups of seven can be chosen to work on a project? d. Suppose two team members insist on either working together or not at all on projects. How many groups of seven can be chosen to work on a project?

1716 ;

700 ;

1715 ;

658 ;

1254 ;

792

Step-by-step explanation:

Given that :

Number of members (n) = 13

a. How many ways can a group of seven be chosen to work on a project?

13C7:

Recall :

nCr = n! ÷ (n-r)! r!

13C7 = 13! ÷ (13 - 7)!7!

= 13! ÷ 6! 7!

(13*12*11*10*9*8*7!) ÷ 7! (6*5*4*3*2*1)

1235520 / 720

= 1716

b. Suppose seven team members are women and six are men.

Men = 6 ; women = 7

(i) How many groups of seven can be chosen that contain four women and three men?

(7C4) * (6C3)

Using calculator :

7C4 = 35

6C3 = 20

(35 * 20) = 700

(ii) How many groups of seven can be chosen that contain at least one man?

13C7 - 7C7

7C7 = only women

13C7 = 1716

7C7 = 1

1716 - 1 = 1715

(iii) How many groups of seven can be chosen that contain at most three women?

(6C4 * 7C3) + (6C5 * 7C2) + (6C6 * 7C1)

Using calculator :

(15 * 35) + (6 * 21) + (1 * 7)

525 + 126 + 7

= 658

c. Suppose two team members refuse to work together on projects. How many groups of seven can be chosen to work on a project?

(First in second out) + (second in first out) + (both out)

13 - 2 = 11

11C6 + 11C6 + 11C7

Using calculator :

462 + 462 + 330

= 1254

d. Suppose two team members insist on either working together or not at all on projects. How many groups of seven can be chosen to work on a project?

Number of ways with both in the group = 11C5

Number of ways with both out of the group = 11C7

11C5 + 11C7

462 + 330

= 792

answer: the answer is e

step-by-step explanation:

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)