Mathematics, 29.07.2019 15:00 bri1814

Carlos’s hockey team has ten players, not counting the goalie. five non goalie players need to be selected for a starting line-up. a)how many different starting lineups are possible if positions are not assigned? b)how many starting lineups are possible if positions are assigned?

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Answer from: aphillips2004

There are 10 players to choose from and you need to choose 5.

The difference between part A and part B is that in A positions are NOT assigned so every person, regardless of when they are chosen has the same role. In part B they have different roles (positions).

For a second let's present there are 2 people (Jay and Jo) and we need to choose 2 of them. If we don't assign them a position there is only one way to do this, you have to take both people Jay and Jo. Let's assume now that there are positions (offense and defense)...now there are 2 ways to pick them...Jay for offense and Jo for defense or Jay for defense and Jo for offense. The order in which we pick them matters in the second scenario because each pick means a particular position. We also get more options this way.

Mathematically in A you have a situation that can be modeled with a combination and in B with a permutation.

PART A
Use the combination nCr as order does not matter. This is a fraction. $10C5= \frac{10!}{5!(10-5)!}$

In symbols you have: $nCr= \frac{n!}{r!(n-r)!}$ .
Recall, n! = (n)(n-1)(n-2)(n-3)...(1)

In this case you get: $10C5= \frac{10!}{5!(10-5)!}$
$= \frac{10!}{5!5!}$
$\frac{(10)(9)(8)(7)(6)(5)(4)(3)(2)(1)}{(5)(4)(3)(2)(1)(5)(4)(3)(2)(1)}$
Any number appearing in the numerator and denominator can be crossed off (divided out) so you end up with:
$\frac{(10)(9)(8)(7)(6)}{(5)(4)(3)(2)(1)} = 252$

PART B
Here the order matters so we use a permutation. Specifically, nPr which we can evaluate this way (start with n and count down r terms). Here we obtain: 10P5=(10)(9)(8)(7)(6)=151,200

Answer from: Quest

x= 1

step-by-step explanation: convert 5\frac{11}{12}5

to improper fraction. use this rule: a \frac{b}{c}=\frac{ac+b}{c}a

ac+b

3.25+0.5x+1\frac{1}{3}+\frac{5}{6}x=\frac{5\times 12+11}{12}3.25+0.5x+1

5×12+11

2 simplify 5\times 125×12 to 6060

3.25+0.5x+1\frac{1}{3}+\frac{5}{6}x=\frac{60+11}{12}3.25+0.5x+1

60+11

3 simplify 60+1160+11 to 7171

3.25+0.5x+1\frac{1}{3}+\frac{5}{6}x=\frac{71}{12}3.25+0.5x+1

4 simplify \frac{5}{6}x

x to \frac{5x}{6}

3.25+0.5x+1\frac{1}{3}+\frac{5x}{6}=\frac{71}{12}3.25+0.5x+1

5 convert 1\frac{1}{3}1

to improper fraction. use this rule: a \frac{b}{c}=\frac{ac+b}{c}a

ac+b

3.25+0.5x+\frac{1\times 3+1}{3}+\frac{5x}{6}=\frac{71}{12}3.25+0.5x+

1×3+1

6 simplify 1\times 31×3 to 33

3.25+0.5x+\frac{3+1}{3}+\frac{5x}{6}=\frac{71}{12}3.25+0.5x+

3+1

7 simplify 3+13+1 to 44

3.25+0.5x+\frac{4}{3}+\frac{5x}{6}=\frac{71}{12}3.25+0.5x+

8 simplify 3.25+0.5x+\frac{4}{3}+\frac{5x}{6}3.25+0.5x+

to \frac{13.75}{3}+\frac{4x}{3}

13.75

\frac{13.75}{3}+\frac{4x}{3}=\frac{71}{12}

13.75

9 simplify \frac{13.75}{3}

13.75

to 4.5833334.583333

4.583333+\frac{4x}{3}=\frac{71}{12}4.583333+

10 subtract 4.5833334.583333 from both sides

\frac{4x}{3}=\frac{71}{12}-4.583333

−4.583333

11 simplify \frac{71}{12}-4.583333

−4.583333 to \frac{4}{3}

\frac{4x}{3}=\frac{4}{3}

12 multiply both sides by 33

4x=\frac{4}{3}\times 34x=

×3

13 simplify \frac{4}{3}\times 3

×3 to \frac{12}{3}

4x=\frac{12}{3}4x=

14 simplify \frac{12}{3}

to 44

4x=44x=4

15 divide both sides by 44

x=1

Answer from: Quest

the answer is 3/2 and for decimal: 1.5

hope i

-lavira: )

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