A starting lineup in basketball consists of two guards, two forwards, and a center.

(a) A certain college team has on its roster 3 centers, 4 guards, 5 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.
(b) Now suppose the roster has 4 guards, 4 forwards, 4 centers, and 2 "swing players" (X and Y) who can play either guard or forward. If 5 of the 14 players are randomly selected, what is the probability that they constitute a legitimate starting lineup?

a. 390 ways

b. 0.82

Explanation

Given

Number of guard = 2

Forward = 2

Center = 1

a.

The college team has on its roster 3 centers, 4 guards, 5 forwards, and one individual (X)

Considering the lineups without X, then lineups with X as guard, then lineups with X as forward.

Note that order doesn't matter

Calculating the lineup without player X:

Selection of 1 Center from 3 center = 3C1 = 3

Selection of 2 forwards from 5 forwards = 5C2 = 10

And

Selection of 2 guards from 4 = 4C2 = 6

Total = 6 * 10 * 3 = 180

X as a guard: (only one player is needed since X has already been selected)

Selection of 1 Center from 3 center = 3C1 = 3

Selection of 2 forwards from 5 forwards = 5C2 = 10

And

Selection of 1 guards from 4 = 4C1= 4

Total = 4 * 10 * 3 = 120

X as a forward: (only one player is needed since X has already been selected)

Selection of 1 Center from 3 center = 3C1 = 3

Selection of 1 forwards from 5 forwards = 5C1 = 5

And

Selection of 2 guards from 4 = 4C2= 6

Total = 6 * 5 * 3 = 90

Total Selection = 90 + 120 + 180 = 390 ways

b.

Considering that player X and Y sit out, player X play while Y sits out, player Y plays while X sits out and player X and Y play (either as a guard, forward or split duties).

Calculating the lineup without player X and Y

Selection of 1 Center from 4 center = 4C1 = 4

Selection of 2 forwards from 4 forwards = 4C2 = 6

And

Selection of 2 guards from 4 = 4C2 = 6

Total = 4 * 6 * 6 = 144

Calculating the lineup without player X but not Y

X as a forward: (only one player is needed since X has already been selected)

If X plays as forward

Selection of 1 Center from 4 center = 4C1 = 4

X plays as forward = 4C1 = 4

And

Selection of 2 guards from 4 = 4C2= 6

Total = 4 * 4 * 6 = 96

If X plays as guard

Selection of 1 Center from 4 = 4C1 = 4

Selection of 2 forwards from 4 = 4C2 = 6

And

X as guard = 4C1= 4

Total = 4 * 6 * 4 = 96

So, if X play and Y sits out; total = 96 + 96 = 192

It's same if Y plays and X sits out = 192

Considering that both X and Y plays

As guards

Selection of 1 Center from 4 = 4C1 = 4

Selection of 2 forward from 4 = 4C2 = 6

And

X and Y as guard = 4C0 = 1

Total = 4 * 6 * 1 = 24

As forwards

Selection of 1 Center from 4 = 4C1 = 4

Selection of 2 guards from 4 = 4C2 = 6

And

X and Y as forward = 4C0 = 1

Total = 4 * 6 * 1 = 24

Both on split duties

Selection of 1 Center from 4 = 4C1 = 4

Selection of1 from 4 forward= 4C1 = 4

And

One from 4 forward = 4C1= 4

Total = 4 * 4 * 4 = 72

Total = 72 +24 + 24 = 120

Total Required Selection = 144+192+192+120 = 648

Overall Possible Selection = 12C5 = 792

Probability = 648/792 = 0.82 Approximated

you answer is a

step-by-step explanation:

the correct answer to your question is a. 4^4

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