1. ## probability question

question 1)
3 friends and 17 other boys are to be randomly divided into 5 teams of 4 players each.what is the probability that the 3 friends find themselves separated,each playing for a different team?

question 2)
an ordinary deck of 52 cards is shuffled.what is the probability that the top 4 cards have
a)different denominations?
b)different suits?
for ques 2,why cant i using the choosing method?

2. 1) the 1st friend can go in any of the 5 teams, 2nd one can go in any of the rest 4 and 3rd friend can go in any of the rest 3 teams. The other 17 players can occupy the rest 17 places in 17! ways. On the other hand, without any restriction, the 20 people could be allotted in 20! ways.
You should be able to find out the probability now.

2) Where are you stuck at? And what does "choosing method" mean?

3. Originally Posted by Sambit
1) the 1st friend can go in any of the 5 teams, 2nd one can go in any of the rest 4 and 3rd friend can go in any of the rest 3 teams. The other 17 players can occupy the rest 17 places in 17! ways. On the other hand, without any restriction, the 20 people could be allotted in 20! ways. You should be able to find out the probability now.
@Sambit: did you know that there are $\displaystyle \dfrac{20!}{(4!)^5\cdot 5!}$ ways to form five teams of four each?

4. Originally Posted by Sambit
1)

2) Where are you stuck at? And what does "choosing method" mean?
for part a), my answer is (13C1)(12C1)(11C1)(10C1)4!/(52C4)
and thx for question 1!

5. Hello, violette!

Question 1
3 friends and 17 other boys are to be randomly divided into 5 teams of 4 players each.
What is the probability that the 3 friends find themselves separated,
each playing for a different team?

Visualize the five teams . . . call them $\displaystyle A,B,C,D,E.$

. . $\displaystyle \begin{array}{|c|c|c|c|c|} \hline \\[-5mm] A & B & C & D & E \\ **** & **** & **** & **** & **** \\ \hline \end{array}$

Friend #1 can be any of the 20 players:
. . $\displaystyle P(\text{friend \#1 is on a team}) \,=\,\frac{20}{20}\,=\,1$

Suppose friend #1 is on Team A.
Then friend #2 can be any of the 16 players on the other four teams.
. . $\displaystyle P(\text{friend \#2 is on a different team}) \,=\,\frac{16}{19}$

Suppose friend #2 is on Team B.
Then friend #3 can be any of the 12 players on the other three teams.
. . $\displaystyle P(\text{friend \#3 is on a different team}) \,=\,\frac{12}{18}\,=\,\frac{2}{3}$

$\displaystyle \displaystyle \text{Therefore: }\:P(\text{3 friends on different teams}) \:=\:1\cdot\frac{16}{19}\cdot\frac{2}{3} \:=\:\frac{32}{57}$

6. Originally Posted by Soroban
Hello, violette!

Visualize the five teams . . . call them $\displaystyle A,B,C,D,E.$

. . $\displaystyle \begin{array}{|c|c|c|c|c|} \hline \\[-5mm]$$\displaystyle A & B & C & D & E \\ **** & **** & **** & **** & **** \\ \hline \end{array}$

Friend #1 can be any of the 20 players:
. . $\displaystyle P(\text{friend \#1 is on a team}) \,=\,\frac{20}{20}\,=\,1$

Suppose friend #1 is on Team A.
Then friend #2 can be any of the 16 players on the other four teams.
. . $\displaystyle P(\text{friend \#2 is on a different team}) \,=\,\frac{16}{19}$

Suppose friend #2 is on Team B.
Then friend #3 can be any of the 12 players on the other three teams.
. . $\displaystyle P(\text{friend \#3 is on a different team}) \,=\,\frac{12}{18}\,=\,\frac{2}{3}$

$\displaystyle \displaystyle \text{Therefore: }\:P(\text{3 friends on different teams}) \:=\:1\cdot\frac{16}{19}\cdot\frac{2}{3} \:=\:\frac{32}{57}$
$\displaystyle \displaystyle \text{Therefore: }\:P(\text{3 friends on different teams}) \:=\:1\cdot\frac{16}{19}\cdot\frac{2}{3} \:=\:\frac{32}{57}$
@ Plato: should the answer be $\displaystyle \frac{(5*4*3*17!)/((3!^2)*(4!^2)*5!)}{\frac{20!}{(4!)^5\cdot 5!}}$?