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The group consists of 8 Indians and 10 Americans, and 10 members will be chosen. How will the members be chosen so that there are at least 5 Americans and at most 4 Indians?

Option: 1

22761


Option: 2

22760


Option: 3

23761

 


Option: 4

23760


Answers (1)

best_answer

Given that,

The group consists of 8 Indians and 10 Americans.

Case I: 6 Americans and 4 Indians can be selected 

The number of ways is given by,

\begin{aligned} & { }^{10} C_6 \times{ }^8 C_4=\frac{10 !}{6 ! 4 !} \times \frac{8 !}{4 ! 4 !} \\ & { }^{10} C_6 \times{ }^8 C_4=210 \times 70 \\ & { }^{10} C_6 \times{ }^8 C_4=14700 \end{aligned}

 

Case II: 7 Americans and 3 Indians can be selected 

The number of ways is given by,

\begin{aligned} &{ }^{10} C_7 \times{ }^8 C_3=\frac{10 !}{7 ! 3 !} \times \frac{8 !}{3 ! 5 !}\\ &{ }^{10} C_7 \times{ }^8 C_3=120 \times 56 \end{aligned}

\begin{aligned} &{ }^{10} C_7 \times{ }^8 C_3=6720 \end{aligned}

Case III: 8 Americans and 2 Indians can be selected 

The number of ways is given by,

\begin{aligned} &{ }^{10} C_8 \times{ }^8 C_2=\frac{10 !}{8 ! 3 !} \times \frac{8 !}{2 ! 6 !}\\ &{ }^{10} C_8 \times{ }^8 C_2=45 \times 28 \end{aligned}

\begin{aligned} &{ }^{10} C_8 \times{ }^8 C_2=1260 \end{aligned}

Case IV: 9 Americans and 1 Indian can be selected 

The number of ways is given by,

\begin{aligned} & { }^{10} C_9 \times{ }^8 C_1=\frac{10 !}{9 ! 1 !} \times \frac{8 !}{1 ! 7 !} \\ & { }^{10} C_9 \times{ }^8 C_1=10 \times 8 \\ & { }^{10} C_9 \times{ }^8 C_1=80 \end{aligned}

Case V: 10 Americans are selected 

The number of ways is given by,

\begin{aligned} & { }^{10} C_6 \times{ }^8 C_4=\frac{10 !}{6 ! 4 !} \times \frac{8 !}{4 ! 4 !} \\ & { }^{10} C_{10}=\frac{10 !}{10 ! 0 !} \\ & { }^{10} C_{10}=1 \end{aligned}

Therefore, the total number of ways of selecting the members = 14700 + 6720 + 1260 + 80+1 = 22761 ways.

 

Posted by

Irshad Anwar

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