Answer
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Hint: In this given question, there are three sections. According to the given sections hints are also given section wise.
Section1: In this section exactly three girls are selected in the committee. So if three girls are selected then only four boys can be selected from the group of nine boys for that particular committee of seven members. So, first we choose 3 girls among 4 girls and 4 boys among 9 boys.
Section2: In this section, at least 3 girls are selected in this committee so it is possible that the committee can have 3 girls or 4 girls. So, if we select 3 girls then we have to select 4 boys and if we select 4 girls then we have to select 3 boys.
Section3: In this section, we are given the condition that at most 3 girls can be selected to form the committee of 7 members. So it is possible that from 0 to 3 girls can be selected for the committee. According to that if 0 girls are selected then 7 boys will be selected. If 1 girl is selected then 6 boys will be selected. If 2 girls are selected then 5 boys will be selected. Similarly, if 3 girls are selected then 4 boys will be selected.
Complete step by step solution:
There three sections of his question in each step we will solve each section separately.
Step1: According to section 1, the committee has exactly 3 girls and the rest are boys. So, we have to choose 4 boys among 9 boys. So,
Numbers of way the committee will be formed are
\[\begin{array}{l}
= 4{C_3} + 9{C_4}\\
= 4 \times 126 = 504
\end{array}\]
Hence, numbers of ways selecting the committee are 504
Step2: In section 2, at least 3 girls will be selected so possibilities of boys will be 3 or 4. So,
\[\begin{array}{l}
= 4{C_3} \times 9{C_4} + 4{C_4} \times 9{C_3}\\
= 4 \times 126 + 1 \times 84\\
= 504 + 84\\
= 588
\end{array}\]
Hence the numbers of ways to form the committee are 588
Step3: In section 3, at most 3 girls will be selected. So, the possibilities will be
\[\begin{array}{l}
0girls7boys = 4{C_0} \times 9{C_7} = 36\\
1girls6boys = 4{C_1} \times 9{C_6} = 336\\
2girls5boys = 4{C_2} \times 9{C_5} = 756\\
3girls4boys = 4{C_3} \times 9{C_4} = 126\\
\end{array}\]
Numbers of ways to form the committee are
\[\begin{array}{l}
= (0girls + 7boys) + (1girls + 6boys) + (2girls + 5boys) + (3girls + 4boys)\\
= 36 + 336 + 756 + 126\\
= 1254\\
\end{array}\]
Note: In this question, three sections of questions have different concepts of selecting the girls and boys. So, while dealing with each section we should give more emphasis on the given conditions such as ‘at least’, ‘at most’ and terms such as ‘exactly’.
Section1: In this section exactly three girls are selected in the committee. So if three girls are selected then only four boys can be selected from the group of nine boys for that particular committee of seven members. So, first we choose 3 girls among 4 girls and 4 boys among 9 boys.
Section2: In this section, at least 3 girls are selected in this committee so it is possible that the committee can have 3 girls or 4 girls. So, if we select 3 girls then we have to select 4 boys and if we select 4 girls then we have to select 3 boys.
Section3: In this section, we are given the condition that at most 3 girls can be selected to form the committee of 7 members. So it is possible that from 0 to 3 girls can be selected for the committee. According to that if 0 girls are selected then 7 boys will be selected. If 1 girl is selected then 6 boys will be selected. If 2 girls are selected then 5 boys will be selected. Similarly, if 3 girls are selected then 4 boys will be selected.
Complete step by step solution:
There three sections of his question in each step we will solve each section separately.
Step1: According to section 1, the committee has exactly 3 girls and the rest are boys. So, we have to choose 4 boys among 9 boys. So,
Numbers of way the committee will be formed are
\[\begin{array}{l}
= 4{C_3} + 9{C_4}\\
= 4 \times 126 = 504
\end{array}\]
Hence, numbers of ways selecting the committee are 504
Step2: In section 2, at least 3 girls will be selected so possibilities of boys will be 3 or 4. So,
\[\begin{array}{l}
= 4{C_3} \times 9{C_4} + 4{C_4} \times 9{C_3}\\
= 4 \times 126 + 1 \times 84\\
= 504 + 84\\
= 588
\end{array}\]
Hence the numbers of ways to form the committee are 588
Step3: In section 3, at most 3 girls will be selected. So, the possibilities will be
\[\begin{array}{l}
0girls7boys = 4{C_0} \times 9{C_7} = 36\\
1girls6boys = 4{C_1} \times 9{C_6} = 336\\
2girls5boys = 4{C_2} \times 9{C_5} = 756\\
3girls4boys = 4{C_3} \times 9{C_4} = 126\\
\end{array}\]
Numbers of ways to form the committee are
\[\begin{array}{l}
= (0girls + 7boys) + (1girls + 6boys) + (2girls + 5boys) + (3girls + 4boys)\\
= 36 + 336 + 756 + 126\\
= 1254\\
\end{array}\]
Note: In this question, three sections of questions have different concepts of selecting the girls and boys. So, while dealing with each section we should give more emphasis on the given conditions such as ‘at least’, ‘at most’ and terms such as ‘exactly’.
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