Subsets and permutations line
There are 20 members of a basketball team.
(a)
The coach must select 12 players to travel to an away game. How many ways are there to select the players who will travel?
(b)
From the 12 players who will travel, the coach must select her starting line-up. She will select a player for each of the five positions: center, power forward, small forward, shooting guard and point guard. How many ways are there for her to select the starting line-up?
(c)
From the 12 players who will travel, the coach must select her starting line-up. She will select a player for each of the five positions: center, power forward, small forward, shooting guard and point guard. However, there are only three of the 12 players who can play center. Otherwise, there are no restrictions. How many ways are there for her to select the starting line-up?
For the (a) i have 20 chose 12 and (b) P(12,5).
However, for the (c) my logic was the following :
So we have 5 sports and the one on the middle is taken by 1 of 3 players. So since there is no order (single seat), we have have 3 choices.
Then, we have 4 seats left and (12-1) 11 players, so it's P(11,4)
Final : 3 * P(11,4)
Is it ok? thanks!
probability
asked Nov 27 '18 at 2:09
Laura1999
202
add a comment |
There are 20 members of a basketball team.
(a)
The coach must select 12 players to travel to an away game. How many ways are there to select the players who will travel?
(b)
From the 12 players who will travel, the coach must select her starting line-up. She will select a player for each of the five positions: center, power forward, small forward, shooting guard and point guard. How many ways are there for her to select the starting line-up?
(c)
From the 12 players who will travel, the coach must select her starting line-up. She will select a player for each of the five positions: center, power forward, small forward, shooting guard and point guard. However, there are only three of the 12 players who can play center. Otherwise, there are no restrictions. How many ways are there for her to select the starting line-up?
For the (a) i have 20 chose 12 and (b) P(12,5).
However, for the (c) my logic was the following :
So we have 5 sports and the one on the middle is taken by 1 of 3 players. So since there is no order (single seat), we have have 3 choices.
Then, we have 4 seats left and (12-1) 11 players, so it's P(11,4)
Final : 3 * P(11,4)
Is it ok? thanks!
probability
asked Nov 27 '18 at 2:09
Laura1999
202
1
Looks OK to me.
– Gerry Myerson
Nov 27 '18 at 3:16
add a comment |
There are 20 members of a basketball team.
(a)
The coach must select 12 players to travel to an away game. How many ways are there to select the players who will travel?
(b)
From the 12 players who will travel, the coach must select her starting line-up. She will select a player for each of the five positions: center, power forward, small forward, shooting guard and point guard. How many ways are there for her to select the starting line-up?
(c)
From the 12 players who will travel, the coach must select her starting line-up. She will select a player for each of the five positions: center, power forward, small forward, shooting guard and point guard. However, there are only three of the 12 players who can play center. Otherwise, there are no restrictions. How many ways are there for her to select the starting line-up?
For the (a) i have 20 chose 12 and (b) P(12,5).
However, for the (c) my logic was the following :
So we have 5 sports and the one on the middle is taken by 1 of 3 players. So since there is no order (single seat), we have have 3 choices.
Then, we have 4 seats left and (12-1) 11 players, so it's P(11,4)
Final : 3 * P(11,4)
Is it ok? thanks!
probability
asked Nov 27 '18 at 2:09
Laura1999
202
There are 20 members of a basketball team.
(a)
The coach must select 12 players to travel to an away game. How many ways are there to select the players who will travel?
(b)
From the 12 players who will travel, the coach must select her starting line-up. She will select a player for each of the five positions: center, power forward, small forward, shooting guard and point guard. How many ways are there for her to select the starting line-up?
(c)
From the 12 players who will travel, the coach must select her starting line-up. She will select a player for each of the five positions: center, power forward, small forward, shooting guard and point guard. However, there are only three of the 12 players who can play center. Otherwise, there are no restrictions. How many ways are there for her to select the starting line-up?
For the (a) i have 20 chose 12 and (b) P(12,5).
However, for the (c) my logic was the following :
So we have 5 sports and the one on the middle is taken by 1 of 3 players. So since there is no order (single seat), we have have 3 choices.
Then, we have 4 seats left and (12-1) 11 players, so it's P(11,4)
Final : 3 * P(11,4)
Is it ok? thanks!
probability
probability
asked Nov 27 '18 at 2:09
Laura1999
202
asked Nov 27 '18 at 2:09
Laura1999
202
asked Nov 27 '18 at 2:09
Laura1999
202
asked Nov 27 '18 at 2:09
Laura1999
202
asked Nov 27 '18 at 2:09
Laura1999
202
202
1
Looks OK to me.
– Gerry Myerson
Nov 27 '18 at 3:16
add a comment |
1
Looks OK to me.
– Gerry Myerson
Nov 27 '18 at 3:16
1
1
Looks OK to me.
– Gerry Myerson
Nov 27 '18 at 3:16
Looks OK to me.
– Gerry Myerson
Nov 27 '18 at 3:16
add a comment |
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1
Looks OK to me.
– Gerry Myerson
Nov 27 '18 at 3:16