You have A,B,C,D,E.
Tie A,B,C together, then you have 3 items you can arrange in 6 ways.
But A,B,C can be arranged 6 ways among themselves.
6*6=36
Five friends, A,B,C,D and D, stand in a line for a photograph.
1) How many different possible arrangements are there if A , B and C stand next to each other?
2) How many different possible arrangements are there if none of A, B and C stand next to each other?
3) if all possible arrangements are equally likely, find the probability that two of A, B and C are next to each other, but the third is not next to either of the other two.
Hello, badi6!
galactus is absolutely correct . . .Five friends, , stand in a line for a photograph.
1) How many different possible arrangements are there
if stand next to each other?
Tie together and there are three "people" to arrange: .
. . There are: . ways.
But can be ordered in ways.
. . Therefore, there are: . ways.
Place D and E in positions #2 and #4: .2) How many different possible arrangements are there
if none of stand next to each other?
. . Then there are ways to place
But D and E can be placed like this: .
. . And there are more ways to place
Therefore, there are: . ways.
I have a laundry-list approach to this one . . .3) If all possible arrangements are equally likely,
find the probability that two of are next to each other,
but the third is not next to either of the other two.
There are three cases:
. . are together and is nonadjacent.
. . are together and is nonadjacent.
. . are together and is nonadjacent.
Case (1): are together and is nonadjacent.
Then there are four scenarios to consider . . .
can be reversed (two choices).
can be in 3 or 4 (two choices).
Then and can be placed (two choices).
. . There are: . ways.
can be reversed (two choices).
must be in 4 (one choice).
Then and can be placed (2 choices).
. . There are: . ways.
can be reversed (two choices).
must be in 1 (one choice).
Then and can be placed (two choices).
. . There are: . ways.
can be reversed (two choices).
can be in 1 or 2 (two choices).
Then and can be placed (2 choices).
. . There are: . ways.
Hence, there are: . ways
. . to have together and nonadjacent.
Since there are 3 cases, there are: . ways
. . to have two of together and the third nonadjacent.
Since there are possible arrangements,
. . the probability is: .
many thanks for your help!
Plz could you also hep as much as you could to tackle the quetion number 5!
http://www.robinson.homechoice.co.uk...on%20Paper.pdf
Thanks