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Mrac [35]
3 years ago
11

Suppose we have a group of 10 men and 15 women.

Mathematics
1 answer:
Law Incorporation [45]3 years ago
7 0

Answer:

a) 15!×16×15×...×7

b) \binom{15}{3}\binom{10}{3}\binom{19}{2}.

c)  25\cdot 24\cdot \binom{23}{4}\binom{19}{5}\binom{14}{3}\binom{11}{3}\binom{9}{3}

Step-by-step explanation:

a) There are 15! ways to line the women first. Now, a man can be placed either to the left of some woman (giving 15 choices) or to the right of the last woman on the line (1 additional choice). So the number of ways of placing the men is 16×15×...×7, that is, 10-permutations of 16.  Then the number of ways to line everyone is 15!×16×15×...×7.

b) First, choose 3 women of said committee. There are 15 women and we must choose 3, then there are \binom{15}{3} ways of doing this. Afterwards, choose 3 men for the commitee. There are 10 men, so the number of ways of doing this is  \binom{10}{3}. The commitee must have 8 people and we already chose 6, so we must choose 2 more persons. There are 25 people in total and we chose 6 of them before, so there are 19 who aren't in the committee. Then the number of ways of completing the commitee is \binom{19}{2}. The number of ways of doing all of this (and creating the commitee) is \binom{15}{3}\binom{10}{3}\binom{19}{2}.

c) There are 25 choices for a president. After choosing a president, there are 24 choices for the secretary. 23 people remain, then there are  \binom{23}{4} ways of choosing the social commitee. We are left with 19 people, thus there are \binom{19}{5} ways to form the social commitee. Similarly, because 14 people haven't been selected, there are  \binom{14}{3}\binom{11}{3}\binom{9}{3} ways to form the working groups. We conclude that there are 25\cdot 24\cdot \binom{23}{4}\binom{19}{5}\binom{14}{3}\binom{11}{3}\binom{9}{3} ways of conforming all the groups.

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Which graphs below have a domain of all real numbers? Which have a range of all real numbers? Which are functions?
slava [35]

Answer:

Step-by-step explanation:

From the picture attached,

Graph (A)

Domain of the function : (-∞, ∞) or a set of real numbers

Range of the function : (-∞, ∞) or a set of real numbers

Graph (B),

Domain : (-∞, ∞) Or a set of real numbers

Range : (-∞, 6] Or x ≤ 6

Graph (C),

Domain : [-2, ∞) or x ≥ -2

Range : (-∞, ∞) Or a set of real numbers

Therefore, Graphs (A), (B) have the domain of all real numbers and Graph (C) has the range of all real numbers.

Since all the parabolas are not the functions,

Therefore, Graph (C) is not a function. Graphs (A) and (B) are the functions.

3 0
3 years ago
The sum of the first three terms of geometric sequence is 14. If the first term is 2, find the possible values of the sum of the
Irina18 [472]

Geometric series are in the form of

a +a*r +a*r^2+...

Where a is the first term and r is the common ratio .

And it is given that

2 +2*r +2*r^2=14

2r^2+2r-12=0

r^2 +r-6=0

(r+3)(r-2)=0

r=-3,2

So the first five terms are

2+2(-3)+2(-3)^2+2(-3)^3+2(-3)^4  or 2+2(2)+2(2)^2+2(2)^3+2(2)^4

= 2-6+18-54+162 or 2+4+8+16+32

= 122 or 62

4 0
2 years ago
Read 2 more answers
Add. (3b + 5) + (2b + 4)
zaharov [31]

Answer:

\huge{ \fbox{ \sf{5b + 9}}}

Step-by-step explanation:

\star{ \text{( \: 3b + 5) + (2b + 4)}}

\text{Step \: 1 \:  :  \sf{Remove \: the \: unnecessary \: parentheses}}

\text{When \: there \: is \: a \: ( + ) \: in \: front \: of \: an \: expression \: in \: parentheses \:, there \: is \: no \: need \: to \: change \: the \: sign \: of \: each \: term \: in \: the \: expression.}

\text{That \: means, \: the \: expression \: remains \: the \: same. \: Just \: you \: have \: to \: remove \: the \: parentheses.}

\mapsto{ \sf{3b + 5 + 2b + 4}}

\text{Step \: 2 \:  : Collect \: like \: terms.}

\text{Like \: terms \: are \: those \: which \: have \: the \: same \: base.}

\mapsto{ \sf{3b + 2b + 5 + 4}}

\mapsto{ \sf{5b + 5 + 4}}

\text{ Step \: 3 \:  :  \sf{Add \: the \: numbers : 5 \: and \: 4}}

\mapsto{ \boxed {\sf{ {5b + 9}}}}

Hope I helped!

Best regards! :D

~\sf{TheAnimeGirl}

4 0
3 years ago
if a town with a population of 500 doubles in size ever 9 years, what will the population be 36 years from now?
s344n2d4d5 [400]

The population of the town in 36 years would be  8000.

<h3>What would be the population of the town in 36 years?</h3>

The formula that can be used to determine the town's population is:
FV = P (1 + r)^n

Where:

  • FV = Future value
  • P = Present value
  • R = rate of growth = 100%
  • N = number of years = 36/9 = 4

500 x 2^4 = 8000

To learn more about future value, please check: brainly.com/question/18760477

4 0
2 years ago
What is the simplified expression for 4 power 4 multiplied by 4 power 3 over 4 power 5? (1 point)
raketka [301]

The solution to the problem is 16.

first, you must set up the equation.

4^4*\frac{4^3}{4^5}

Then, simplify exponents.

256*\frac{64}{1024}

Next, simplify the equation.

16

5 0
3 years ago
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