A. C(13, 10) = 13! = 13·12·11 = 13 · 2 · 11 = 286.C(13, 10) = 13! = 13·12·11 = 13 · 2 · 11 = 286.
B. P(13,10)= 13! =13! =13·12·11·10·9·8·7·6·5·4.
(13−10)! 3!
C. f there is exactly one woman chosen, this is possible in C(10, 9)C(3, 1) =
10! 3!
9!1! 1!2!
10! 3!
8!2! 2!1!
10! 3!
7!3! 3!0!
= 10 · 3 = 30 ways; two women chosen — in C(10,8)C(3,2) =
= 45·3 = 135 ways; three women chosen — in C(10, 7)C(3, 3) =
= 10·9·8 ·1 = 120 ways. Altogether there are 30+135+120 = 285
1·2·3
<span>possible choices.</span><span>
</span>
Answer:
its 125
Step-by-step explanation:
5x5x5
Answer:
Step-by-step explanation:
Answer:
See Below.
Step-by-step explanation:
Paragraph Proof:
We are given that ∠1 ≅ ∠4. ∠1 and ∠4 are alternate exterior angles. Since they are congruent, by the Alternate Exterior Angles Converse, the two lines being cut by the transversal must be parallel. Then by the Alternate Interior Angles Theorem, ∠3 ≅ ∠2.
2-Column Proof
Statements Reasons:
1) 
Given
2) 
Alternate Exterior Angles Converse
3) 
Alternate Interior Angles Theorem
I would write a flowchart as well, but unfortunately, I never learned it that way.
