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In how many ways ways can the coach make substitutions in a soccer match?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)How many ways he can attempt the paper?how many ways to arrange 10 players between 3 positions with at least two players in each position?Combinatorics - how many ways for 10 people to make a line with restrictionsNumber of ways in which 6 rings can be worn on the 4 fingers of one handHow many numbers of $7$ digits can be formed with the digit $0,1,1,5,6,6,6$.How many different ways can the numbers 1-9 be arranged in a 3x9 grid?Where is my solution wrong for this combinatorics problem?How many strings of $6$ digits are there which use only the digits $0, 1$, or $2$ and in which $2$, whenever it appears, always does so after $1$?A hotel has 6 single rooms, 6 double rooms and 4 rooms in which 3 persons can stay. In how many ways can 30 persons be accommodated in this hotel?In how many ways can the people line up if the first woman appears ahead of the first child?
$begingroup$
My solution:
Coach can pick any one of the 11 subs and then place the sub in any one of the 11 places so that's (11)(11) ways
Then coach can pick any of the 10 remaining subs and place them again in 11 places in (10)(11) ways
Finally coach can pick any of the 9 remaining subs and place them in any of the 11 places 9(11) ways
Therefore total is (11)(11)× (10)(11)×(9)(11) +1 way for no subs = 1317691
Divided by 1000 the remaining amt is 691
combinatorics
edited Apr 2 at 8:53
N. F. Taussig
45.5k103358
asked Mar 31 at 3:02
user122343user122343
1076
$endgroup$
add a comment |
$begingroup$
My solution:
Coach can pick any one of the 11 subs and then place the sub in any one of the 11 places so that's (11)(11) ways
Then coach can pick any of the 10 remaining subs and place them again in 11 places in (10)(11) ways
Finally coach can pick any of the 9 remaining subs and place them in any of the 11 places 9(11) ways
Therefore total is (11)(11)× (10)(11)×(9)(11) +1 way for no subs = 1317691
Divided by 1000 the remaining amt is 691
combinatorics
edited Apr 2 at 8:53
N. F. Taussig
45.5k103358
asked Mar 31 at 3:02
user122343user122343
1076
$endgroup$
2
$begingroup$
You've counted the possibilities for three substitutions or none, but there could also be one or two.
$endgroup$
– FredH
Mar 31 at 3:11
$begingroup$
Ahh yes I forgot to do that
$endgroup$
– user122343
Mar 31 at 3:16
1
$begingroup$
I'm guessing for two substitutions it would be (11)(11)×(10)(11) and for one is (11)(11) . And after adding together and dividing I got 122
$endgroup$
– user122343
Mar 31 at 3:18
add a comment |
$begingroup$
My solution:
Coach can pick any one of the 11 subs and then place the sub in any one of the 11 places so that's (11)(11) ways
Then coach can pick any of the 10 remaining subs and place them again in 11 places in (10)(11) ways
Finally coach can pick any of the 9 remaining subs and place them in any of the 11 places 9(11) ways
Therefore total is (11)(11)× (10)(11)×(9)(11) +1 way for no subs = 1317691
Divided by 1000 the remaining amt is 691
combinatorics
edited Apr 2 at 8:53
N. F. Taussig
45.5k103358
asked Mar 31 at 3:02
user122343user122343
1076
$endgroup$
My solution:
Coach can pick any one of the 11 subs and then place the sub in any one of the 11 places so that's (11)(11) ways
Then coach can pick any of the 10 remaining subs and place them again in 11 places in (10)(11) ways
Finally coach can pick any of the 9 remaining subs and place them in any of the 11 places 9(11) ways
Therefore total is (11)(11)× (10)(11)×(9)(11) +1 way for no subs = 1317691
Divided by 1000 the remaining amt is 691
combinatorics
combinatorics
edited Apr 2 at 8:53
N. F. Taussig
45.5k103358
asked Mar 31 at 3:02
user122343user122343
1076
edited Apr 2 at 8:53
N. F. Taussig
45.5k103358
asked Mar 31 at 3:02
user122343user122343
1076
edited Apr 2 at 8:53
N. F. Taussig
45.5k103358
edited Apr 2 at 8:53
N. F. Taussig
45.5k103358
edited Apr 2 at 8:53
N. F. Taussig
45.5k103358
45.5k103358
asked Mar 31 at 3:02
user122343user122343
1076
asked Mar 31 at 3:02
user122343user122343
1076
asked Mar 31 at 3:02
user122343user122343
1076
1076
2
$begingroup$
You've counted the possibilities for three substitutions or none, but there could also be one or two.
$endgroup$
– FredH
Mar 31 at 3:11
$begingroup$
Ahh yes I forgot to do that
$endgroup$
– user122343
Mar 31 at 3:16
1
$begingroup$
I'm guessing for two substitutions it would be (11)(11)×(10)(11) and for one is (11)(11) . And after adding together and dividing I got 122
$endgroup$
– user122343
Mar 31 at 3:18
add a comment |
2
$begingroup$
You've counted the possibilities for three substitutions or none, but there could also be one or two.
$endgroup$
– FredH
Mar 31 at 3:11
$begingroup$
Ahh yes I forgot to do that
$endgroup$
– user122343
Mar 31 at 3:16
1
$begingroup$
I'm guessing for two substitutions it would be (11)(11)×(10)(11) and for one is (11)(11) . And after adding together and dividing I got 122
$endgroup$
– user122343
Mar 31 at 3:18
2
2
$begingroup$
You've counted the possibilities for three substitutions or none, but there could also be one or two.
$endgroup$
– FredH
Mar 31 at 3:11
$begingroup$
You've counted the possibilities for three substitutions or none, but there could also be one or two.
$endgroup$
– FredH
Mar 31 at 3:11
$begingroup$
Ahh yes I forgot to do that
$endgroup$
– user122343
Mar 31 at 3:16
$begingroup$
Ahh yes I forgot to do that
$endgroup$
– user122343
Mar 31 at 3:16
1
1
$begingroup$
I'm guessing for two substitutions it would be (11)(11)×(10)(11) and for one is (11)(11) . And after adding together and dividing I got 122
$endgroup$
– user122343
Mar 31 at 3:18
$begingroup$
I'm guessing for two substitutions it would be (11)(11)×(10)(11) and for one is (11)(11) . And after adding together and dividing I got 122
$endgroup$
– user122343
Mar 31 at 3:18
add a comment |
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2
$begingroup$
You've counted the possibilities for three substitutions or none, but there could also be one or two.
$endgroup$
– FredH
Mar 31 at 3:11
$begingroup$
Ahh yes I forgot to do that
$endgroup$
– user122343
Mar 31 at 3:16
1
$begingroup$
I'm guessing for two substitutions it would be (11)(11)×(10)(11) and for one is (11)(11) . And after adding together and dividing I got 122
$endgroup$
– user122343
Mar 31 at 3:18