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3 years ago

Given a group of 8 women and 11 men, how many different ways are there of choosing one man and one woman for a committee?

1 answer:

8 0

Any of the 11 man can be chosen, and combined with any of the 8 women.

Assume we select man1. The selected committee can be:

(m1,w1), (m1,w2), (m1,w3), (m1,w4), (m1,w5), (m1,w6), (m1,w7), (m1,w8),

so there are 8 committees selections with man1 in them.

we could repeat the same procedure for the remaining 10 men, and get 8 committees where each of them is a member.

so there are 11*8=88 ways of choosing 1 man and 1 woman.

Answer: 88

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2 years ago

The junior and senior students at Mathville High School are going to present an exciting musical entitled, "Math, What is it Goo

xeze [42]

Some parts are missing in the queston. Find attached the picture with the complete question

Answer:

$\large\boxed{\large\boxed{161}}$

Explanation:

Let's put the information in a table step-by step.

(number of remaining students)

Juniors Seniors

Condition

Initially J S

15 seniors left S - 15

Twice juniors as seniors 2(S - 15)

3/4 of the juniors left 1/4×2(S - 15)
1/3 of seniors left 2/3×(S - 15)

At the end, there were 8 more seniors than juniors:

2/3×(S - 15) - 1/4×2(S - 15) = 8

Now you have obtained one equation, which you can solve to find S, the number of senior students, and then the number of junior students.

Solve the equation:

$2/3\times (S - 15) - 1/4\times 2(S - 15) = 8$

Mutilply all by 12:

$8(S - 15)-6(S - 15)=96$

Distribution property:

$8S-120-6S-90=96$

Addtion property of equalities:

$8S-6S=96+120+90$

Add like terms:

$2S=306$

Division property of equalities:

$S=306/2=153$

That is the number of senior students that came out to the information meeting, but the number of students remaining to perform in the school musical is (from the table above):

$2/3\times (S-15)+1/4\times 2(S-15)$