Probability : permutation or choose
A club has $18$ members.
(a) How many ways are there to choose four members of the club to serve on an executive committee?
Solution: $binom{18}{4} = 3060$
(b) How many ways are there to choose a president, vice president, secretary, and treasurer of the club, where no person can hold more than one office?
Solution: $P(18,4)=18 cdot 17 cdot 16 cdot 15=73,440$
Sometimes, I am confused when to pick up the permutation (P) or the choose. Correct me if i am wrong but,
When the order matters $to$ choose
When the order does not matter $to$ permutation
combinatorics discrete-mathematics
edited Nov 29 at 9:40
N. F. Taussig
43.5k93355
asked Nov 28 at 23:57
Laura1999
202
add a comment |
A club has $18$ members.
(a) How many ways are there to choose four members of the club to serve on an executive committee?
Solution: $binom{18}{4} = 3060$
(b) How many ways are there to choose a president, vice president, secretary, and treasurer of the club, where no person can hold more than one office?
Solution: $P(18,4)=18 cdot 17 cdot 16 cdot 15=73,440$
Sometimes, I am confused when to pick up the permutation (P) or the choose. Correct me if i am wrong but,
When the order matters $to$ choose
When the order does not matter $to$ permutation
combinatorics discrete-mathematics
edited Nov 29 at 9:40
N. F. Taussig
43.5k93355
asked Nov 28 at 23:57
Laura1999
202
2
Other way 'round and you're good!
– munchhausen
Nov 29 at 0:00
1
Go for understanding, not slogans.
– Gerry Myerson
Nov 29 at 1:00
Type$binom{n}{k}$to obtain $binom{n}{k}$. Please read this tutorial on how to typeset mathematics on this site. Also, you are not calculating a probability here. Problems of enumeration should be tagged combinatorics.
– N. F. Taussig
Nov 29 at 9:38
add a comment |
A club has $18$ members.
(a) How many ways are there to choose four members of the club to serve on an executive committee?
Solution: $binom{18}{4} = 3060$
(b) How many ways are there to choose a president, vice president, secretary, and treasurer of the club, where no person can hold more than one office?
Solution: $P(18,4)=18 cdot 17 cdot 16 cdot 15=73,440$
Sometimes, I am confused when to pick up the permutation (P) or the choose. Correct me if i am wrong but,
When the order matters $to$ choose
When the order does not matter $to$ permutation
combinatorics discrete-mathematics
edited Nov 29 at 9:40
N. F. Taussig
43.5k93355
asked Nov 28 at 23:57
Laura1999
202
A club has $18$ members.
(a) How many ways are there to choose four members of the club to serve on an executive committee?
Solution: $binom{18}{4} = 3060$
(b) How many ways are there to choose a president, vice president, secretary, and treasurer of the club, where no person can hold more than one office?
Solution: $P(18,4)=18 cdot 17 cdot 16 cdot 15=73,440$
Sometimes, I am confused when to pick up the permutation (P) or the choose. Correct me if i am wrong but,
When the order matters $to$ choose
When the order does not matter $to$ permutation
combinatorics discrete-mathematics
combinatorics discrete-mathematics
edited Nov 29 at 9:40
N. F. Taussig
43.5k93355
asked Nov 28 at 23:57
Laura1999
202
edited Nov 29 at 9:40
N. F. Taussig
43.5k93355
asked Nov 28 at 23:57
Laura1999
202
edited Nov 29 at 9:40
N. F. Taussig
43.5k93355
edited Nov 29 at 9:40
N. F. Taussig
43.5k93355
edited Nov 29 at 9:40
N. F. Taussig
43.5k93355
43.5k93355
asked Nov 28 at 23:57
Laura1999
202
asked Nov 28 at 23:57
Laura1999
202
asked Nov 28 at 23:57
Laura1999
202
202
2
Other way 'round and you're good!
– munchhausen
Nov 29 at 0:00
1
Go for understanding, not slogans.
– Gerry Myerson
Nov 29 at 1:00
Type$binom{n}{k}$to obtain $binom{n}{k}$. Please read this tutorial on how to typeset mathematics on this site. Also, you are not calculating a probability here. Problems of enumeration should be tagged combinatorics.
– N. F. Taussig
Nov 29 at 9:38
add a comment |
2
Other way 'round and you're good!
– munchhausen
Nov 29 at 0:00
1
Go for understanding, not slogans.
– Gerry Myerson
Nov 29 at 1:00
Type$binom{n}{k}$to obtain $binom{n}{k}$. Please read this tutorial on how to typeset mathematics on this site. Also, you are not calculating a probability here. Problems of enumeration should be tagged combinatorics.
– N. F. Taussig
Nov 29 at 9:38
2
2
Other way 'round and you're good!
– munchhausen
Nov 29 at 0:00
Other way 'round and you're good!
– munchhausen
Nov 29 at 0:00
1
1
Go for understanding, not slogans.
– Gerry Myerson
Nov 29 at 1:00
Go for understanding, not slogans.
– Gerry Myerson
Nov 29 at 1:00
Type
$binom{n}{k}$ to obtain $binom{n}{k}$. Please read this tutorial on how to typeset mathematics on this site. Also, you are not calculating a probability here. Problems of enumeration should be tagged combinatorics.– N. F. Taussig
Nov 29 at 9:38
Type
$binom{n}{k}$ to obtain $binom{n}{k}$. Please read this tutorial on how to typeset mathematics on this site. Also, you are not calculating a probability here. Problems of enumeration should be tagged combinatorics.– N. F. Taussig
Nov 29 at 9:38
add a comment |
1 Answer
1
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oldest
votes
(a) How many ways are there to choose four members of the club to serve on an executive committee?
Solution: $binom {18}4=3060$
You are selecting four from eigteen members.
If you are counting ways to select items, that is a count of combinations. Combinations are not distinguished by ordering. (Order is not important.)
$binom n r$ or $^nmathrm C_r$ counts combinations of $r$ elements selected from a set of $n$
$${^nmathrm C_r}= dfrac{n!}{r!cdot (n-r)!}$$
(b) How many ways are there to choose a president, vice president, secretary, and treasurer of the club, where no person can hold more than one office?
Solution: $P(18,4)=18⋅17⋅16⋅15=73,440$
You are selecting and arranging four from eighteen members into four specific positions.
If you are counting ways to select and arrange items, that is a count of permutations. Permutations are distinguished by ordering. (Order is important!)
$P(n,r)$ or $^nmathrm P_r$ counts permutations of arrangements for $r$ elements selected from a set of $n$.
$${^nmathrm P_r}= dfrac{n!}{(n-r)!}$$
${^nmathrm P_r}={^nmathrm C_r}cdot r!$
answered Nov 29 at 3:08
Graham Kemp
84.7k43378
add a comment |
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votes
(a) How many ways are there to choose four members of the club to serve on an executive committee?
Solution: $binom {18}4=3060$
You are selecting four from eigteen members.
If you are counting ways to select items, that is a count of combinations. Combinations are not distinguished by ordering. (Order is not important.)
$binom n r$ or $^nmathrm C_r$ counts combinations of $r$ elements selected from a set of $n$
$${^nmathrm C_r}= dfrac{n!}{r!cdot (n-r)!}$$
(b) How many ways are there to choose a president, vice president, secretary, and treasurer of the club, where no person can hold more than one office?
Solution: $P(18,4)=18⋅17⋅16⋅15=73,440$
You are selecting and arranging four from eighteen members into four specific positions.
If you are counting ways to select and arrange items, that is a count of permutations. Permutations are distinguished by ordering. (Order is important!)
$P(n,r)$ or $^nmathrm P_r$ counts permutations of arrangements for $r$ elements selected from a set of $n$.
$${^nmathrm P_r}= dfrac{n!}{(n-r)!}$$
${^nmathrm P_r}={^nmathrm C_r}cdot r!$
answered Nov 29 at 3:08
Graham Kemp
84.7k43378
add a comment |
(a) How many ways are there to choose four members of the club to serve on an executive committee?
Solution: $binom {18}4=3060$
You are selecting four from eigteen members.
If you are counting ways to select items, that is a count of combinations. Combinations are not distinguished by ordering. (Order is not important.)
$binom n r$ or $^nmathrm C_r$ counts combinations of $r$ elements selected from a set of $n$
$${^nmathrm C_r}= dfrac{n!}{r!cdot (n-r)!}$$
(b) How many ways are there to choose a president, vice president, secretary, and treasurer of the club, where no person can hold more than one office?
Solution: $P(18,4)=18⋅17⋅16⋅15=73,440$
You are selecting and arranging four from eighteen members into four specific positions.
If you are counting ways to select and arrange items, that is a count of permutations. Permutations are distinguished by ordering. (Order is important!)
$P(n,r)$ or $^nmathrm P_r$ counts permutations of arrangements for $r$ elements selected from a set of $n$.
$${^nmathrm P_r}= dfrac{n!}{(n-r)!}$$
${^nmathrm P_r}={^nmathrm C_r}cdot r!$
answered Nov 29 at 3:08
Graham Kemp
84.7k43378
add a comment |
(a) How many ways are there to choose four members of the club to serve on an executive committee?
Solution: $binom {18}4=3060$
You are selecting four from eigteen members.
If you are counting ways to select items, that is a count of combinations. Combinations are not distinguished by ordering. (Order is not important.)
$binom n r$ or $^nmathrm C_r$ counts combinations of $r$ elements selected from a set of $n$
$${^nmathrm C_r}= dfrac{n!}{r!cdot (n-r)!}$$
(b) How many ways are there to choose a president, vice president, secretary, and treasurer of the club, where no person can hold more than one office?
Solution: $P(18,4)=18⋅17⋅16⋅15=73,440$
You are selecting and arranging four from eighteen members into four specific positions.
If you are counting ways to select and arrange items, that is a count of permutations. Permutations are distinguished by ordering. (Order is important!)
$P(n,r)$ or $^nmathrm P_r$ counts permutations of arrangements for $r$ elements selected from a set of $n$.
$${^nmathrm P_r}= dfrac{n!}{(n-r)!}$$
${^nmathrm P_r}={^nmathrm C_r}cdot r!$
answered Nov 29 at 3:08
Graham Kemp
84.7k43378
(a) How many ways are there to choose four members of the club to serve on an executive committee?
Solution: $binom {18}4=3060$
You are selecting four from eigteen members.
If you are counting ways to select items, that is a count of combinations. Combinations are not distinguished by ordering. (Order is not important.)
$binom n r$ or $^nmathrm C_r$ counts combinations of $r$ elements selected from a set of $n$
$${^nmathrm C_r}= dfrac{n!}{r!cdot (n-r)!}$$
(b) How many ways are there to choose a president, vice president, secretary, and treasurer of the club, where no person can hold more than one office?
Solution: $P(18,4)=18⋅17⋅16⋅15=73,440$
You are selecting and arranging four from eighteen members into four specific positions.
If you are counting ways to select and arrange items, that is a count of permutations. Permutations are distinguished by ordering. (Order is important!)
$P(n,r)$ or $^nmathrm P_r$ counts permutations of arrangements for $r$ elements selected from a set of $n$.
$${^nmathrm P_r}= dfrac{n!}{(n-r)!}$$
${^nmathrm P_r}={^nmathrm C_r}cdot r!$
answered Nov 29 at 3:08
Graham Kemp
84.7k43378
edited Nov 29 at 3:20
answered Nov 29 at 3:08
Graham Kemp
84.7k43378
answered Nov 29 at 3:08
Graham Kemp
84.7k43378
answered Nov 29 at 3:08
Graham Kemp
84.7k43378
84.7k43378
add a comment |
add a comment |
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2
Other way 'round and you're good!
– munchhausen
Nov 29 at 0:00
1
Go for understanding, not slogans.
– Gerry Myerson
Nov 29 at 1:00
Type
$binom{n}{k}$to obtain $binom{n}{k}$. Please read this tutorial on how to typeset mathematics on this site. Also, you are not calculating a probability here. Problems of enumeration should be tagged combinatorics.– N. F. Taussig
Nov 29 at 9:38