Mathematics, 01.08.2019 12:30 gaby8282

Astarting lineup in basketball consists of two guards, two forwards, and a center. (a) a certain college team has on its roster four centers, five guards, three forwards, and one individual (x) who can play either guard or forward. how many different starting lineups can be created? [hint: consider lineups without x, then lineups with x as guard, then lineups with x as forward.] 240 lineups

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Answer from: allieballey0727

1 . Consider lineups without x. You have to select

2 guards from 5 quards;2 forwards from 3 forwards;1 center from 4 centers.

It can be made in

$C_5^2\cdot C_3^2\cdot C_4^1=10\cdot 3\cdot 4=120$ different ways.

2. Consider lineups with x as guard. You have to select

2 guards from 6 quards (one of them must be x);2 forwards from 3 forwards;1 center from 4 centers.

It can be made in

$C_5^1\cdot C_3^2\cdot C_4^1=5\cdot 3\cdot 4=60$ different ways.

3. Consider lineups with x as forward. You have to select

2 guards from 5 quards;2 forwards from 4 forwards (one of them must be x);1 center from 4 centers.

It can be made in

$C_5^2\cdot C_3^1\cdot C_4^1=10\cdot 3\cdot 4=120$ different ways.

Therefore, the total number of different lineups is

120+60+120=300.

Answer from: stuckonquestions

300

Justification:

Follow the hint:

1) Line ups without x:

Number of guards to pick: 2 of 5 => C5,2
Number of forwards to pick: 2 of 3 => C3,2
Number of centers to pick 1 of 4 => C4,1

Number of possible combinations: C5,2 * C3,2 * C4,1

The formula for Cm,n is: Cm,n = m! / [n! * (m-n)! ]

=>

C5,2 = 5! / [2! (5-3)! ] = 5*4 / 2 = 10

C3,2 = 3! / [2! (3-1! ] = 3

C4,1 = 4! / [1!(4-1)!] = 4

=> 10*3*4 = 120 different starting lineups without x

2) Lineups with x as guard

You only need to pick one more guard

number of guards to pick: 1 of 5 => C5,1
number of forwards to pick: 2 of 3 => C3,2
number of centers to pick: 4 => 1 of 4 => C4,1

Number of possible combinations: C5,1 * C3,2 * C4,1 = 5 * 3 * 4 = 60

3) Lineups with x as a forward,

You only need to pick one more forward:

Number of guards to pick: 2 of 5 => C5,2
Number of forwards to pick: 1 of 3 => C3,1
Number of centers to pick: 1 of 4 => C4,1

.Number of possible combinations: C5,2 * C3,1 * C4,1 = 10 * 3 * 4 = 120

4) Total number of different lineups:

Number of lineups without x + number of lineups with x as guard + number of lineups with x as forward = 120 + 60 + 120 = 300

Answer from: Quest

Hello! mark me brainliest
answer:
congruent segments
step-by-step explanation:
congruent segments. take a line, cut it in half, superipose one half over the other half; you'll have congruence.

Answer from: Quest

well we know there are 4 clubs in 52 cards so it’s replaced so it should be 3/51

step-by-step explanation:

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Astarting lineup in basketball consists of two guards, two forwards, and a center. (a) a certain col...