Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • In how many ways can an Expert team of one or more team leaders be made by dissolving the Group I of 16 different experts,the Group II of 5 different experts, and the Group III of 6 diffe

In how many ways can an Expert team of one or more team leaders be made by dissolving the Group I of 16 different experts,the Group II of 5 different experts, and the Group III of 6 different experts?

Option: 1

715


Option: 2

2^{27}


Option: 3

2^{27}-1


Option: 4

490


Answers (1)

Note the following:

  • The formula for the combination for the selection of the \mathrm{x} items from the \mathrm{y} different items is \mathrm{={ }^y \mathrm{C}_x=\frac{y !}{x !(y-x) !}}
  • The combination for the selection of the one or more items from the \mathrm{{ }^n} different items is \mathrm{ =2^n-1 }

From the available data, the following is evident.

  • The combination for the selection of the one or more different experts from the Group\mathrm{ \text { I is }=\left({ }^4 C_0+{ }^4 C_1+\ldots \ldots \ldots+{ }^{16} C_{16}\right) }
  • The combination for the selection of the one or more different experts from the Group \mathrm{ \text { II is }=\left({ }^5 C_0+{ }^5 C_1+\ldots \ldots+{ }^5 C_5\right) }
  • The combination for the selection of the one or more different experts from the Group \mathrm{ \text { III is }=\left({ }^6 C_0+{ }^5 C_1+\ldots \ldots+{ }^6 C_6\right) }

Therefore, the required combination for selection of one or more team leader is

=\mathrm{ \left({ }^4 C_0+{ }^4 C_1+\ldots \ldots+{ }^16 C_14\right)\left({ }^5 C_0+{ }^ 5C_1+\ldots \ldots+{ }^5 C_5\right)\left({ }^6 C_0+\ldots \ldots+{ }^6 C_6\right)-1 }

=2^{16}\times2^{5}\times2^{6}-1

=2^{27}-1

 

 

 

Posted by

Ramraj Saini

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Latest Question