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lawyer [7]
3 years ago
11

Among the seven nominees for two vacancies on a city council are three men and four women. In how many ways can these vacancies

be filled
Mathematics
1 answer:
irga5000 [103]3 years ago
8 0

This question is incomplete

Complete Question

Among the seven nominees for two vacancies on the city council are three men and four women. In how many ways may these vacancies be filled

a) with any two of the nominees?

b) with any two of the women?

c) with one of the men and one of the women?

Answer:

a) 21 ways

b) 6 ways

c) 12 ways

Step-by-step explanation:

We solve this question using combination formula

C(n, r) = nCr = n!/r! (n - r)!

a) with any two of the nominees?

Probability (two of the nominees) = 7C2

= 7!/2! ×(7 - 2)!

= 7!/ 2! × 5!

= 7 × 6 × 5 × 4 × 3 × 2 × 1/2 × 1 ×(5 × 4 × 3 × 2 × 1)

= 21

b) with any two of the women?

We have a total of 4 women

Hence, the probability of any two of the four women, filling the vacancies =

P(any two of the women) = 4C2

= 4!/2! ×( 4 - 2)!

= 4!/ 2! × 2!

= 4 × 3 × 2 × 1/ 2 × 1 ×( 2 × 1)

= 6

c) with one of the men and one of the

Total number of men = 3

Total number of women = 4

= 3C1 × 4C1

= [3!/1! ×(3 - 1)! ] × [4!/1! ×(4 - 1)! ]

= [3!/1! × 2!] × [4!/1! ×3!]

= [3 × 2 × 1/ 1 × 2 × 1] × [4 × 3 × 2 × 1/ 1 × 3 × 2 × 1]

= 3 × [24/6]

= 3 × 4

= 12 ways

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Step-by-step explanation:

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3 years ago
What is the square root of 245?
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2 years ago
Use mathematical induction to prove the statement is true for all positive integers n, or show why it is false:
kondaur [170]
\text{Proof by induction:}
\text{Test that the statement holds or n = 1}

LHS = (3 - 2)^{2} = 1
RHS = \frac{6 - 4}{2} = \frac{2}{2} = 1 = LHS
\text{Thus, the statement holds for the base case.}

\text{Assume the statement holds for some arbitrary term, n= k}
1^{2} + 4^{2} + 7^{2} + ... + (3k - 2)^{2} = \frac{k(6k^{2} - 3k - 1)}{2}

\text{Prove it is true for n = k + 1}
RTP: 1^{2} + 4^{2} + 7^{2} + ... + [3(k + 1) - 2]^{2} = \frac{(k + 1)[6(k + 1)^{2} - 3(k + 1) - 1]}{2} = \frac{(k + 1)[6k^{2} + 9k + 2]}{2}

LHS = \underbrace{1^{2} + 4^{2} + 7^{2} + ... + (3k - 2)^{2}}_{\frac{k(6k^{2} - 3k - 1)}{2}} + [3(k + 1) - 2]^{2}
= \frac{k(6k^{2} - 3k - 1)}{2} + [3(k + 1) - 2]^{2}
= \frac{k(6k^{2} - 3k - 1) + 2[3(k + 1) - 2]^{2}}{2}
= \frac{k(6k^{2} - 3k - 1) + 2(3k + 1)^{2}}{2}
= \frac{k(6k^{2} - 3k - 1) + 18k^{2} + 12k + 2}{2}
= \frac{k(6k^{2} - 3k - 1 + 18k + 12) + 2}{2}
= \frac{k(6k^{2} + 15k + 11) + 2}{}
= \frac{(k + 1)[6k^{2} + 9k + 2]}{2}
= \frac{(k + 1)[6(k + 1)^{2} - 3(k + 1) - 1]}{2}
= RHS

Since it is true for n = 1, n = k, and n = k + 1, by the principles of mathematical induction, it is true for all positive values of n.
3 0
3 years ago
Kari drew two parallel lines PQ and RS intersected by a transversal KL, as shown below:
Arlecino [84]
Given that PQ and RS are drawn with KL as tranversal intersecting PQ at M and RS at point N. Angle QMN is congruent to angle LNS because they are alternate to each other. The theorem that Kari can use to show that the meansure of QML is supplementary to the measure of angle SNK is Alternate Exterior Angles Theorem.
This is because angle KNR is equal to QML by alternate exterior angles theorem so is angle MLP and SNK 
3 0
3 years ago
Read 2 more answers
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