2 particular adults cannot be together in a selection
$begingroup$
A 4 member committee is to be formed from a group of 9 adults.Find the number of ways this committee can be formed if 2 particular adults cannot be together.
my attempt:
Case 1- N(both not in committee) = 7C4
case 2- N(one in and one out) = 7C3
addition of 2 will give my answer.
However, the mistake in my step is in case 2-
N(one in and one out) = 7C3 X 2
why must i multiply by 2 ?
combinatorics
asked Nov 27 '18 at 14:20
ErikienErikien
494
$endgroup$
add a comment |
$begingroup$
A 4 member committee is to be formed from a group of 9 adults.Find the number of ways this committee can be formed if 2 particular adults cannot be together.
my attempt:
Case 1- N(both not in committee) = 7C4
case 2- N(one in and one out) = 7C3
addition of 2 will give my answer.
However, the mistake in my step is in case 2-
N(one in and one out) = 7C3 X 2
why must i multiply by 2 ?
combinatorics
asked Nov 27 '18 at 14:20
ErikienErikien
494
$endgroup$
add a comment |
$begingroup$
A 4 member committee is to be formed from a group of 9 adults.Find the number of ways this committee can be formed if 2 particular adults cannot be together.
my attempt:
Case 1- N(both not in committee) = 7C4
case 2- N(one in and one out) = 7C3
addition of 2 will give my answer.
However, the mistake in my step is in case 2-
N(one in and one out) = 7C3 X 2
why must i multiply by 2 ?
combinatorics
asked Nov 27 '18 at 14:20
ErikienErikien
494
$endgroup$
A 4 member committee is to be formed from a group of 9 adults.Find the number of ways this committee can be formed if 2 particular adults cannot be together.
my attempt:
Case 1- N(both not in committee) = 7C4
case 2- N(one in and one out) = 7C3
addition of 2 will give my answer.
However, the mistake in my step is in case 2-
N(one in and one out) = 7C3 X 2
why must i multiply by 2 ?
combinatorics
combinatorics
asked Nov 27 '18 at 14:20
ErikienErikien
494
asked Nov 27 '18 at 14:20
ErikienErikien
494
asked Nov 27 '18 at 14:20
ErikienErikien
494
asked Nov 27 '18 at 14:20
ErikienErikien
494
asked Nov 27 '18 at 14:20
ErikienErikien
494
494
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Because there could be two possibilities in case $2$ , for example if person $A$ has been selected in a team then remaining ways would be $7 choose 3$ . However person $B$ can also be selected in the same way , therefore you need to multiply the answer by $2$.
Therefore final answer is $7choose 4$ $+$ $7choose 3$$×2$.
answered Nov 27 '18 at 14:24
Akash RoyAkash Roy
1
$endgroup$
add a comment |
$begingroup$
Person $1$ and person $2$ cannot be together. In your case $2$, you calculated, for instance, person $1$ in and person $2$ out. However, there's the other way around, namely person $2$ in and person $1$ out. That's why you should multiply by $2$.
edited Nov 27 '18 at 14:28
Akash Roy
1
answered Nov 27 '18 at 14:26
fbrunogsfbrunogs
113
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Because there could be two possibilities in case $2$ , for example if person $A$ has been selected in a team then remaining ways would be $7 choose 3$ . However person $B$ can also be selected in the same way , therefore you need to multiply the answer by $2$.
Therefore final answer is $7choose 4$ $+$ $7choose 3$$×2$.
answered Nov 27 '18 at 14:24
Akash RoyAkash Roy
1
$endgroup$
add a comment |
$begingroup$
Because there could be two possibilities in case $2$ , for example if person $A$ has been selected in a team then remaining ways would be $7 choose 3$ . However person $B$ can also be selected in the same way , therefore you need to multiply the answer by $2$.
Therefore final answer is $7choose 4$ $+$ $7choose 3$$×2$.
answered Nov 27 '18 at 14:24
Akash RoyAkash Roy
1
$endgroup$
add a comment |
$begingroup$
Because there could be two possibilities in case $2$ , for example if person $A$ has been selected in a team then remaining ways would be $7 choose 3$ . However person $B$ can also be selected in the same way , therefore you need to multiply the answer by $2$.
Therefore final answer is $7choose 4$ $+$ $7choose 3$$×2$.
answered Nov 27 '18 at 14:24
Akash RoyAkash Roy
1
$endgroup$
Because there could be two possibilities in case $2$ , for example if person $A$ has been selected in a team then remaining ways would be $7 choose 3$ . However person $B$ can also be selected in the same way , therefore you need to multiply the answer by $2$.
Therefore final answer is $7choose 4$ $+$ $7choose 3$$×2$.
answered Nov 27 '18 at 14:24
Akash RoyAkash Roy
1
answered Nov 27 '18 at 14:24
Akash RoyAkash Roy
1
answered Nov 27 '18 at 14:24
Akash RoyAkash Roy
1
answered Nov 27 '18 at 14:24
Akash RoyAkash Roy
1
1
add a comment |
add a comment |
$begingroup$
Person $1$ and person $2$ cannot be together. In your case $2$, you calculated, for instance, person $1$ in and person $2$ out. However, there's the other way around, namely person $2$ in and person $1$ out. That's why you should multiply by $2$.
edited Nov 27 '18 at 14:28
Akash Roy
1
answered Nov 27 '18 at 14:26
fbrunogsfbrunogs
113
$endgroup$
add a comment |
$begingroup$
Person $1$ and person $2$ cannot be together. In your case $2$, you calculated, for instance, person $1$ in and person $2$ out. However, there's the other way around, namely person $2$ in and person $1$ out. That's why you should multiply by $2$.
edited Nov 27 '18 at 14:28
Akash Roy
1
answered Nov 27 '18 at 14:26
fbrunogsfbrunogs
113
$endgroup$
add a comment |
$begingroup$
Person $1$ and person $2$ cannot be together. In your case $2$, you calculated, for instance, person $1$ in and person $2$ out. However, there's the other way around, namely person $2$ in and person $1$ out. That's why you should multiply by $2$.
edited Nov 27 '18 at 14:28
Akash Roy
1
answered Nov 27 '18 at 14:26
fbrunogsfbrunogs
113
$endgroup$
Person $1$ and person $2$ cannot be together. In your case $2$, you calculated, for instance, person $1$ in and person $2$ out. However, there's the other way around, namely person $2$ in and person $1$ out. That's why you should multiply by $2$.
edited Nov 27 '18 at 14:28
Akash Roy
1
answered Nov 27 '18 at 14:26
fbrunogsfbrunogs
113
edited Nov 27 '18 at 14:28
Akash Roy
1
edited Nov 27 '18 at 14:28
Akash Roy
1
edited Nov 27 '18 at 14:28
Akash Roy
1
1
answered Nov 27 '18 at 14:26
fbrunogsfbrunogs
113
answered Nov 27 '18 at 14:26
fbrunogsfbrunogs
113
answered Nov 27 '18 at 14:26
fbrunogsfbrunogs
113
113
add a comment |
add a comment |
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