# Math Help - probability

1. ## probability

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.

2. 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

3. ## probability

thanks a lot

and what about the remaining questions?

Thanks

Five friends, $\{A,B,C,D, E\}$, stand in a line for a photograph.

1) How many different possible arrangements are there
if $A, B, C$ stand next to each other?
galactus is absolutely correct . . .

Tie $A,B,C$ together and there are three "people" to arrange: . $\boxed{ABC}, D, E$
. . There are: . $3!$ ways.

But $A,B,C$ can be ordered in $3!$ ways.

. . Therefore, there are: . $3! \times 3! \:=\:\boxed{36}$ ways.

2) How many different possible arrangements are there
if none of $A, B, C$ stand next to each other?
Place D and E in positions #2 and #4: . $\_\:D\:\_\:E\:\_$
. . Then there are $3!$ ways to place $A,B,C.$

But D and E can be placed like this: . $\_\:E\:\_\:D\:\_$
. . And there are $3!$ more ways to place $A,B,C.$

Therefore, there are: . $3! + 3! \:=\:12$ ways.

3) If all possible arrangements are equally likely,
find the probability that two of $A, B, C$ are next to each other,
but the third is not next to either of the other two.
I have a laundry-list approach to this one . . .

There are three cases:
. . $(1)\;A,B$ are together and $C$ is nonadjacent.
. . $(2)\;B,C$ are together and $A$ is nonadjacent.
. . $(3)\;A,C$ are together and $B$ is nonadjacent.

Case (1): $A,B$ are together and $C$ is nonadjacent.
Then there are four scenarios to consider . . .

$(a)\;\begin{array}{cccc} ^1 & ^2 & ^3 & ^4 \\
\boxed{AB} & \_ & \_ & \_\end{array}$

$A,B$ can be reversed (two choices).
$C$ can be in 3 or 4 (two choices).
Then $D$ and $E$ can be placed (two choices).
. . There are: . $2 \times 2 \times 2 \:=\:8$ ways.

$(b)\;\begin{array}{cccc}^1 & ^2 & ^3 & ^4 \\
\_ & \boxed{AB} & \_ & \_ \end{array}$

$A,B$ can be reversed (two choices).
$C$ must be in 4 (one choice).
Then $D$ and $E$ can be placed (2 choices).
. . There are: . $2 \times 1 \times 2 \:=\:4$ ways.

$(c)\;\begin{array}{cccc}^1 & ^2 & ^3 & ^4 \\
\_ & \_ & \boxed{AB} & \_ \end{array}$

$A,B$ can be reversed (two choices).
$C$ must be in 1 (one choice).
Then $D$ and $E$ can be placed (two choices).
. . There are: . $2 \times 1 \times2 \:=\:4$ ways.

$(d)\;\begin{array}{cccc}^1 & ^2 & ^3 & ^4 \\
\_ & \_ & \_ & \boxed{AB} \end{array}$

$A,B$ can be reversed (two choices).
$C$ can be in 1 or 2 (two choices).
Then $D$ and $E$ can be placed (2 choices).
. . There are: . $2 \times 2 \times 2 \:=\:8$ ways.

Hence, there are: . $8 + 4 + 4 + 8 \:=\:24$ ways
. . to have $A, B$ together and $C$ nonadjacent.

Since there are 3 cases, there are: . $3 \times 24 \:=\:72$ ways
. . to have two of $A,B,C$ together and the third nonadjacent.

Since there are $5!=120$ possible arrangements,
. . the probability is: . $\frac{72}{12} \;=\;\boxed{\frac{3}{5}}$