Answer
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Hint: For solving this first, we will find the total number of ways in which a committee can be formed with the help of the formula of the selection of $r$ items from $n$ distinct items. Then, we will solve for each part to get their correct answers.
Complete step-by-step solution -
Given:
It is given that there are 10 professors and 20 students out of whom a committee of 2 professors and 3 students is to be formed.
Now, before we proceed we should know the total number of ways to select $r$ items from $n$ distinct items is ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ .
Now, for making the committee we have to select 2 professors from 10 different professors and 3 students from 20 different students. Then,
Number of ways in which 2 professors can be selected from 10 professors $={}^{10}{{C}_{2}}$ .
Number of ways in which 3 students can be selected from 20 different students $={}^{20}{{C}_{3}}$ .
Now, the total number of ways in which the committee of 2 professors and 3 students can be formed will be equal to ${}^{10}{{C}_{2}}\times {}^{20}{{C}_{3}}=\dfrac{10\times 9}{2}\times \dfrac{20\times 19\times 18}{2\times 3}=51300$ .
Thus, the total number of ways in which the committee can be formed is 51300.
(i) A particular professor is included.
Now, we have to include one particular professor or in other words, we have to select only one professor from the remaining 9 professors as one professor is pre-selected. And we can select one professor from the remaining nine professors in ${}^{9}{{C}_{1}}=9$ ways.
Now, for making the committee we have to select 3 students from 20 different students, and we can do it in ${}^{20}{{C}_{3}}=\dfrac{20\times 19\times 18}{2\times 3}=1140$ ways.
Now, the total number of ways in which the committee of 2 professors and 3 students can be formed such that a particular professor is included will be equal to ${}^{9}{{C}_{1}}\times {}^{20}{{C}_{3}}=10260$ ways.
Thus, the total number of ways in which the committee of 2 professors and 3 students can be formed such that a particular professor is included will be 10260.
(ii) A particular student is included.
Now, we have to include one particular student or in other words, we have to select only two students from the remaining 19 students as one student is pre-selected. And we can select two students from the remaining 19 students in ${}^{19}{{C}_{2}}=\dfrac{19\times 18}{2}=171$ ways.
Now, for making the committee we have to select 2 professors from 10 different professors, and we can do it in ${}^{10}{{C}_{2}}=\dfrac{10\times 9}{2}=45$ ways.
Now, the total number of ways in which the committee of 2 professors and 3 students can be formed such that a particular student is included will be equal to ${}^{19}{{C}_{2}}\times {}^{10}{{C}_{2}}=7695$ ways.
Thus, the total number of ways in which the committee of 2 professors and 3 students can be formed such that a particular student is included will be 7695.
(ii) A particular student is excluded.
Now, we have to exclude one particular student or in other words, we have to select three students from the remaining 19 students as one student is pre-excluded. And we can select three students from the remaining 19 students in ${}^{19}{{C}_{3}}=\dfrac{19\times 18\times 17}{2\times 3}=969$ ways.
Now, for making the committee we have to select 2 professors from 10 different professors, and we can do it in ${}^{10}{{C}_{2}}=\dfrac{10\times 9}{2}=45$ ways.
Now, the total number of ways in which the committee of 2 professors and 3 students can be formed such that a particular student is excluded will be equal to ${}^{19}{{C}_{3}}\times {}^{10}{{C}_{2}}=43605$ ways.
Thus, the total number of ways in which the committee of 2 professors and 3 students can be formed such that a particular student is excluded will be 43605.
Note: Here, the student should first try to understand what is asked in each part and try to make the situation simple and easy to apply the formula directly to get the correct answer quickly. Moreover, avoid making calculation mistakes while solving.
Complete step-by-step solution -
Given:
It is given that there are 10 professors and 20 students out of whom a committee of 2 professors and 3 students is to be formed.
Now, before we proceed we should know the total number of ways to select $r$ items from $n$ distinct items is ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ .
Now, for making the committee we have to select 2 professors from 10 different professors and 3 students from 20 different students. Then,
Number of ways in which 2 professors can be selected from 10 professors $={}^{10}{{C}_{2}}$ .
Number of ways in which 3 students can be selected from 20 different students $={}^{20}{{C}_{3}}$ .
Now, the total number of ways in which the committee of 2 professors and 3 students can be formed will be equal to ${}^{10}{{C}_{2}}\times {}^{20}{{C}_{3}}=\dfrac{10\times 9}{2}\times \dfrac{20\times 19\times 18}{2\times 3}=51300$ .
Thus, the total number of ways in which the committee can be formed is 51300.
(i) A particular professor is included.
Now, we have to include one particular professor or in other words, we have to select only one professor from the remaining 9 professors as one professor is pre-selected. And we can select one professor from the remaining nine professors in ${}^{9}{{C}_{1}}=9$ ways.
Now, for making the committee we have to select 3 students from 20 different students, and we can do it in ${}^{20}{{C}_{3}}=\dfrac{20\times 19\times 18}{2\times 3}=1140$ ways.
Now, the total number of ways in which the committee of 2 professors and 3 students can be formed such that a particular professor is included will be equal to ${}^{9}{{C}_{1}}\times {}^{20}{{C}_{3}}=10260$ ways.
Thus, the total number of ways in which the committee of 2 professors and 3 students can be formed such that a particular professor is included will be 10260.
(ii) A particular student is included.
Now, we have to include one particular student or in other words, we have to select only two students from the remaining 19 students as one student is pre-selected. And we can select two students from the remaining 19 students in ${}^{19}{{C}_{2}}=\dfrac{19\times 18}{2}=171$ ways.
Now, for making the committee we have to select 2 professors from 10 different professors, and we can do it in ${}^{10}{{C}_{2}}=\dfrac{10\times 9}{2}=45$ ways.
Now, the total number of ways in which the committee of 2 professors and 3 students can be formed such that a particular student is included will be equal to ${}^{19}{{C}_{2}}\times {}^{10}{{C}_{2}}=7695$ ways.
Thus, the total number of ways in which the committee of 2 professors and 3 students can be formed such that a particular student is included will be 7695.
(ii) A particular student is excluded.
Now, we have to exclude one particular student or in other words, we have to select three students from the remaining 19 students as one student is pre-excluded. And we can select three students from the remaining 19 students in ${}^{19}{{C}_{3}}=\dfrac{19\times 18\times 17}{2\times 3}=969$ ways.
Now, for making the committee we have to select 2 professors from 10 different professors, and we can do it in ${}^{10}{{C}_{2}}=\dfrac{10\times 9}{2}=45$ ways.
Now, the total number of ways in which the committee of 2 professors and 3 students can be formed such that a particular student is excluded will be equal to ${}^{19}{{C}_{3}}\times {}^{10}{{C}_{2}}=43605$ ways.
Thus, the total number of ways in which the committee of 2 professors and 3 students can be formed such that a particular student is excluded will be 43605.
Note: Here, the student should first try to understand what is asked in each part and try to make the situation simple and easy to apply the formula directly to get the correct answer quickly. Moreover, avoid making calculation mistakes while solving.
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