Mixing
number of ways $3$ teachers can be invited for a guest lecturer on $6$ days in a school
$begingroup$
In how many ways $3$ teachers can be invited for a guest lecturer on $6$ days in a school so that every teacher is invited at least once (on any given day exactly one teacher is invited)
Try: First Teacher $T_{1}$ has $6$ possiability (In any one of $6$ days)
Second Teacher $T_{2}$ has $5$ Possiability
Third Teacher $T_{3}$ has $4$ possiability
So total $6times 5 times 4 = 120$
But answer given as $540$
Could some help me how to count total ways so the answer is $540$
Thanks in advance
combinatorics
edited Jan 9 at 17:06
user
4,0001627
asked Jan 9 at 9:03
DXTDXT
5,6742630
$endgroup$
add a comment |
$begingroup$
In how many ways $3$ teachers can be invited for a guest lecturer on $6$ days in a school so that every teacher is invited at least once (on any given day exactly one teacher is invited)
Try: First Teacher $T_{1}$ has $6$ possiability (In any one of $6$ days)
Second Teacher $T_{2}$ has $5$ Possiability
Third Teacher $T_{3}$ has $4$ possiability
So total $6times 5 times 4 = 120$
But answer given as $540$
Could some help me how to count total ways so the answer is $540$
Thanks in advance
combinatorics
edited Jan 9 at 17:06
user
4,0001627
asked Jan 9 at 9:03
DXTDXT
5,6742630
$endgroup$
3
$begingroup$
You are only filling 3 days. You need a guest on all 6 days. So, some teachers will need to repeat. There does not appear to be a further restriction so it could be that all visit twice or one comes four times and the others just once.
$endgroup$
– badjohn
Jan 9 at 9:08
add a comment |
$begingroup$
In how many ways $3$ teachers can be invited for a guest lecturer on $6$ days in a school so that every teacher is invited at least once (on any given day exactly one teacher is invited)
Try: First Teacher $T_{1}$ has $6$ possiability (In any one of $6$ days)
Second Teacher $T_{2}$ has $5$ Possiability
Third Teacher $T_{3}$ has $4$ possiability
So total $6times 5 times 4 = 120$
But answer given as $540$
Could some help me how to count total ways so the answer is $540$
Thanks in advance
combinatorics
edited Jan 9 at 17:06
user
4,0001627
asked Jan 9 at 9:03
DXTDXT
5,6742630
$endgroup$
In how many ways $3$ teachers can be invited for a guest lecturer on $6$ days in a school so that every teacher is invited at least once (on any given day exactly one teacher is invited)
Try: First Teacher $T_{1}$ has $6$ possiability (In any one of $6$ days)
Second Teacher $T_{2}$ has $5$ Possiability
Third Teacher $T_{3}$ has $4$ possiability
So total $6times 5 times 4 = 120$
But answer given as $540$
Could some help me how to count total ways so the answer is $540$
Thanks in advance
combinatorics
combinatorics
edited Jan 9 at 17:06
user
4,0001627
asked Jan 9 at 9:03
DXTDXT
5,6742630
edited Jan 9 at 17:06
user
4,0001627
asked Jan 9 at 9:03
DXTDXT
5,6742630
edited Jan 9 at 17:06
user
4,0001627
edited Jan 9 at 17:06
user
4,0001627
edited Jan 9 at 17:06
user
4,0001627
4,0001627
asked Jan 9 at 9:03
DXTDXT
5,6742630
asked Jan 9 at 9:03
DXTDXT
5,6742630
asked Jan 9 at 9:03
DXTDXT
5,6742630
5,6742630
3
$begingroup$
You are only filling 3 days. You need a guest on all 6 days. So, some teachers will need to repeat. There does not appear to be a further restriction so it could be that all visit twice or one comes four times and the others just once.
$endgroup$
– badjohn
Jan 9 at 9:08
add a comment |
3
$begingroup$
You are only filling 3 days. You need a guest on all 6 days. So, some teachers will need to repeat. There does not appear to be a further restriction so it could be that all visit twice or one comes four times and the others just once.
$endgroup$
– badjohn
Jan 9 at 9:08
3
3
$begingroup$
You are only filling 3 days. You need a guest on all 6 days. So, some teachers will need to repeat. There does not appear to be a further restriction so it could be that all visit twice or one comes four times and the others just once.
$endgroup$
– badjohn
Jan 9 at 9:08
$begingroup$
You are only filling 3 days. You need a guest on all 6 days. So, some teachers will need to repeat. There does not appear to be a further restriction so it could be that all visit twice or one comes four times and the others just once.
$endgroup$
– badjohn
Jan 9 at 9:08
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Undesired events:
- A single teacher teaches repeately. $3$ ways to do so.
- Only two teachers. First choose one of the teacher to exclude him/her, and then we avoid two extreme cases where one teacher teaches everyday.
$$3cdot (2^6-2)$$
Hence the undesired outcomes are $$3(2^6-1)$$
Without restriction, we could have $3^6$. Hence
$$3^6-3(2^6-1)=3(3^5-2^6+1)=3(243-64+1)=3(180)=540$$
answered Jan 9 at 9:08
Siong Thye GohSiong Thye Goh
101k1466117
$endgroup$
add a comment |
$begingroup$
The table shows the teacher, the number of visits and the number of ways:
$$begin{array}{c|c|c|c}
T_1&T_2&T_3&text{No of visits}\
hline
1&1&4&{6choose 1}{5choose 1}=color{red}{30}\
1&2&3&{6choose 1}{5choose 2}=color{green}{60}\
1&3&2&{6choose 1}{5choose 3}=color{green}{60}\
1&4&1&{6choose 1}{5choose 4}=color{red}{30}\
2&1&3&{6choose 2}{4choose 1}=color{green}{60}\
2&2&2&{6choose 2}{4choose 2}=color{blue}{90}\
2&3&1&{6choose 2}{4choose 3}=color{green}{60}\
3&1&2&{6choose 3}{3choose 1}=color{green}{60}\
3&2&1&{6choose 3}{3choose 2}=color{green}{60}\
4&1&1&{6choose 4}{2choose 1}=color{red}{30}\
hline
&&&color{red}{90}+color{green}{360}+color{blue}{90}=540.
end{array}$$
answered Jan 9 at 18:19
farruhotafarruhota
20.1k2738
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Undesired events:
- A single teacher teaches repeately. $3$ ways to do so.
- Only two teachers. First choose one of the teacher to exclude him/her, and then we avoid two extreme cases where one teacher teaches everyday.
$$3cdot (2^6-2)$$
Hence the undesired outcomes are $$3(2^6-1)$$
Without restriction, we could have $3^6$. Hence
$$3^6-3(2^6-1)=3(3^5-2^6+1)=3(243-64+1)=3(180)=540$$
answered Jan 9 at 9:08
Siong Thye GohSiong Thye Goh
101k1466117
$endgroup$
add a comment |
$begingroup$
Undesired events:
- A single teacher teaches repeately. $3$ ways to do so.
- Only two teachers. First choose one of the teacher to exclude him/her, and then we avoid two extreme cases where one teacher teaches everyday.
$$3cdot (2^6-2)$$
Hence the undesired outcomes are $$3(2^6-1)$$
Without restriction, we could have $3^6$. Hence
$$3^6-3(2^6-1)=3(3^5-2^6+1)=3(243-64+1)=3(180)=540$$
answered Jan 9 at 9:08
Siong Thye GohSiong Thye Goh
101k1466117
$endgroup$
add a comment |
$begingroup$
Undesired events:
- A single teacher teaches repeately. $3$ ways to do so.
- Only two teachers. First choose one of the teacher to exclude him/her, and then we avoid two extreme cases where one teacher teaches everyday.
$$3cdot (2^6-2)$$
Hence the undesired outcomes are $$3(2^6-1)$$
Without restriction, we could have $3^6$. Hence
$$3^6-3(2^6-1)=3(3^5-2^6+1)=3(243-64+1)=3(180)=540$$
answered Jan 9 at 9:08
Siong Thye GohSiong Thye Goh
101k1466117
$endgroup$
Undesired events:
- A single teacher teaches repeately. $3$ ways to do so.
- Only two teachers. First choose one of the teacher to exclude him/her, and then we avoid two extreme cases where one teacher teaches everyday.
$$3cdot (2^6-2)$$
Hence the undesired outcomes are $$3(2^6-1)$$
Without restriction, we could have $3^6$. Hence
$$3^6-3(2^6-1)=3(3^5-2^6+1)=3(243-64+1)=3(180)=540$$
answered Jan 9 at 9:08
Siong Thye GohSiong Thye Goh
101k1466117
answered Jan 9 at 9:08
Siong Thye GohSiong Thye Goh
101k1466117
answered Jan 9 at 9:08
Siong Thye GohSiong Thye Goh
101k1466117
answered Jan 9 at 9:08
Siong Thye GohSiong Thye Goh
101k1466117
101k1466117
add a comment |
add a comment |
$begingroup$
The table shows the teacher, the number of visits and the number of ways:
$$begin{array}{c|c|c|c}
T_1&T_2&T_3&text{No of visits}\
hline
1&1&4&{6choose 1}{5choose 1}=color{red}{30}\
1&2&3&{6choose 1}{5choose 2}=color{green}{60}\
1&3&2&{6choose 1}{5choose 3}=color{green}{60}\
1&4&1&{6choose 1}{5choose 4}=color{red}{30}\
2&1&3&{6choose 2}{4choose 1}=color{green}{60}\
2&2&2&{6choose 2}{4choose 2}=color{blue}{90}\
2&3&1&{6choose 2}{4choose 3}=color{green}{60}\
3&1&2&{6choose 3}{3choose 1}=color{green}{60}\
3&2&1&{6choose 3}{3choose 2}=color{green}{60}\
4&1&1&{6choose 4}{2choose 1}=color{red}{30}\
hline
&&&color{red}{90}+color{green}{360}+color{blue}{90}=540.
end{array}$$
answered Jan 9 at 18:19
farruhotafarruhota
20.1k2738
$endgroup$
add a comment |
$begingroup$
The table shows the teacher, the number of visits and the number of ways:
$$begin{array}{c|c|c|c}
T_1&T_2&T_3&text{No of visits}\
hline
1&1&4&{6choose 1}{5choose 1}=color{red}{30}\
1&2&3&{6choose 1}{5choose 2}=color{green}{60}\
1&3&2&{6choose 1}{5choose 3}=color{green}{60}\
1&4&1&{6choose 1}{5choose 4}=color{red}{30}\
2&1&3&{6choose 2}{4choose 1}=color{green}{60}\
2&2&2&{6choose 2}{4choose 2}=color{blue}{90}\
2&3&1&{6choose 2}{4choose 3}=color{green}{60}\
3&1&2&{6choose 3}{3choose 1}=color{green}{60}\
3&2&1&{6choose 3}{3choose 2}=color{green}{60}\
4&1&1&{6choose 4}{2choose 1}=color{red}{30}\
hline
&&&color{red}{90}+color{green}{360}+color{blue}{90}=540.
end{array}$$
answered Jan 9 at 18:19
farruhotafarruhota
20.1k2738
$endgroup$
add a comment |
$begingroup$
The table shows the teacher, the number of visits and the number of ways:
$$begin{array}{c|c|c|c}
T_1&T_2&T_3&text{No of visits}\
hline
1&1&4&{6choose 1}{5choose 1}=color{red}{30}\
1&2&3&{6choose 1}{5choose 2}=color{green}{60}\
1&3&2&{6choose 1}{5choose 3}=color{green}{60}\
1&4&1&{6choose 1}{5choose 4}=color{red}{30}\
2&1&3&{6choose 2}{4choose 1}=color{green}{60}\
2&2&2&{6choose 2}{4choose 2}=color{blue}{90}\
2&3&1&{6choose 2}{4choose 3}=color{green}{60}\
3&1&2&{6choose 3}{3choose 1}=color{green}{60}\
3&2&1&{6choose 3}{3choose 2}=color{green}{60}\
4&1&1&{6choose 4}{2choose 1}=color{red}{30}\
hline
&&&color{red}{90}+color{green}{360}+color{blue}{90}=540.
end{array}$$
answered Jan 9 at 18:19
farruhotafarruhota
20.1k2738
$endgroup$
The table shows the teacher, the number of visits and the number of ways:
$$begin{array}{c|c|c|c}
T_1&T_2&T_3&text{No of visits}\
hline
1&1&4&{6choose 1}{5choose 1}=color{red}{30}\
1&2&3&{6choose 1}{5choose 2}=color{green}{60}\
1&3&2&{6choose 1}{5choose 3}=color{green}{60}\
1&4&1&{6choose 1}{5choose 4}=color{red}{30}\
2&1&3&{6choose 2}{4choose 1}=color{green}{60}\
2&2&2&{6choose 2}{4choose 2}=color{blue}{90}\
2&3&1&{6choose 2}{4choose 3}=color{green}{60}\
3&1&2&{6choose 3}{3choose 1}=color{green}{60}\
3&2&1&{6choose 3}{3choose 2}=color{green}{60}\
4&1&1&{6choose 4}{2choose 1}=color{red}{30}\
hline
&&&color{red}{90}+color{green}{360}+color{blue}{90}=540.
end{array}$$
answered Jan 9 at 18:19
farruhotafarruhota
20.1k2738
answered Jan 9 at 18:19
farruhotafarruhota
20.1k2738
answered Jan 9 at 18:19
farruhotafarruhota
20.1k2738
answered Jan 9 at 18:19
farruhotafarruhota
20.1k2738
20.1k2738
add a comment |
add a comment |
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3
$begingroup$
You are only filling 3 days. You need a guest on all 6 days. So, some teachers will need to repeat. There does not appear to be a further restriction so it could be that all visit twice or one comes four times and the others just once.
$endgroup$
– badjohn
Jan 9 at 9:08