In how many ways \[{\text{3}}\] teachers can be invited for a guest lecture in \[6\]days in a school so that every teacher is invited at least once?(on any given day exactly one teacher is invited)
A.\[620\]
B.\[540\]
C.\[520\]
D.\[5040\]
Answer
662.1k+ views
Hint: We need to make each and every possible case and then we have to calculate ways for each case. And finally if cases are all unique add them it will be our answer or if some cases are added twice or counted twice so subtract them in order to get a solution without fail.
Complete step-by-step answer:
Given that three teachers has to be invited for a guest lecture in \[6\]days in a school so that every teacher is invited at least once
Now, forming cases in a way that a single teacher can teach repeatedly
So there are \[^{\text{3}}{{\text{C}}_{\text{1}}}{\text{ = 3}}\]ways.
Now, let deal with only two teachers, first choose one of the teacher to exclude the teacher, and then we avoid two
extreme cases in which one teacher teaches every day.
So, the number of ways are \[^{\text{3}}{{\text{C}}_{\text{1}}}{\text{(}}{{\text{2}}^{\text{6}}}{\text{ - 2) = 3(}}{{\text{2}}^{\text{6}}}{\text{ - 2)}}\]
Hence, now we need to add both the cases as a single teacher teaching case and two teachers teaching cases.
So, the number of ways are , \[{\text{3(}}{{\text{2}}^{\text{6}}}{\text{ - 2) - 3 = 3(}}{{\text{2}}^{\text{6}}}{\text{ - 1)}}\]
And now the total number of cases without any restriction are \[{{\text{3}}^6}\].
Hence, we have to subtract total possible without restriction cases from the restricted cases so,
\[{\text{ = }}{{\text{3}}^{\text{6}}}{\text{ - 3(}}{{\text{2}}^{\text{6}}}{\text{ - 1)}}\]
On taking 3 common we get,
\[{\text{ = 3(}}{{\text{3}}^{\text{5}}}{\text{ - }}{{\text{2}}^{\text{6}}}{\text{ + 1)}}\]
On simplification we get,
\[{\text{ = 3(243 - 64 + 1)}}\]
\[{\text{ = 3(180)}}\]
\[{\text{ = 540}}\]
Hence, option (b) is the correct answer.
Note: Permutations are the different ways in which a collection of items can be arranged through various different methods. For example: The different ways in which the alphabets A, B and C can be grouped together, in ways such as ABC, ACB, BCA, CBA, CAB, BAC.
Complete step-by-step answer:
Given that three teachers has to be invited for a guest lecture in \[6\]days in a school so that every teacher is invited at least once
Now, forming cases in a way that a single teacher can teach repeatedly
So there are \[^{\text{3}}{{\text{C}}_{\text{1}}}{\text{ = 3}}\]ways.
Now, let deal with only two teachers, first choose one of the teacher to exclude the teacher, and then we avoid two
extreme cases in which one teacher teaches every day.
So, the number of ways are \[^{\text{3}}{{\text{C}}_{\text{1}}}{\text{(}}{{\text{2}}^{\text{6}}}{\text{ - 2) = 3(}}{{\text{2}}^{\text{6}}}{\text{ - 2)}}\]
Hence, now we need to add both the cases as a single teacher teaching case and two teachers teaching cases.
So, the number of ways are , \[{\text{3(}}{{\text{2}}^{\text{6}}}{\text{ - 2) - 3 = 3(}}{{\text{2}}^{\text{6}}}{\text{ - 1)}}\]
And now the total number of cases without any restriction are \[{{\text{3}}^6}\].
Hence, we have to subtract total possible without restriction cases from the restricted cases so,
\[{\text{ = }}{{\text{3}}^{\text{6}}}{\text{ - 3(}}{{\text{2}}^{\text{6}}}{\text{ - 1)}}\]
On taking 3 common we get,
\[{\text{ = 3(}}{{\text{3}}^{\text{5}}}{\text{ - }}{{\text{2}}^{\text{6}}}{\text{ + 1)}}\]
On simplification we get,
\[{\text{ = 3(243 - 64 + 1)}}\]
\[{\text{ = 3(180)}}\]
\[{\text{ = 540}}\]
Hence, option (b) is the correct answer.
Note: Permutations are the different ways in which a collection of items can be arranged through various different methods. For example: The different ways in which the alphabets A, B and C can be grouped together, in ways such as ABC, ACB, BCA, CBA, CAB, BAC.
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