A planning committee for a major development project must consist six members. There are eight architects and five engineers available to choose from.3.2) In how many ways can a planning committee be chosen with at least three engineers (rounded off to zero decimals)? There is at least one architect and one engineer on the planning committee.

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Asked Sep 26, 2019
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A planning committee for a major development project must consist six members. There are eight architects and five engineers available to choose from.

3.2) In how many ways can a planning committee be chosen with at least three engineers (rounded off to zero decimals)? There is at least one architect and one engineer on the planning committee.

 

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Step 1

Combination:

The number of different selections of ‘r’ elements from a set of ‘n’ elements is denoted and given by,

 

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п! п ПС, (or) (п-rг)! х г!

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Step 2

It is given that the selected committee has six members in it.

There are 8 architects and 5 engineers.

A committee with at least three engineers can have remaining 3 members who are architects or engineers.

Thus, the number of ways a committee be chosen with at least three engineers and remaining architects is as follows:

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Number of ways of selecting Selecting 3 engineers and 3 architects or at least 3 engineers Selecting 4 engineers and 2 architects or for a committe 6 memebers Selecting 5 engineers and 1 architects (5C)(8C)(5C,)(8C,)+(5C,) (8C;) =(10x56)+(5x 28)+ (1x8) = 560+140 + 8 = 708

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Step 3

Thus, the number of ways a committee be chosen with at least three engineers and remaining architects is 708 ways.

Simila...

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Selecting 1 engineer and 5 architects or Number of ways of selecting Selecting 2 engineers and 4 architects or at least 1 engineer and 1 architect Selecting 3 engineers and 3 architects or for a committe 6 memebers Selecting 4 engineers and 2 architects or Selecting 5 engineers and 1 architects (5C)(8C,)(5C) (8C,) + (5C,)(8C;)- (5C,)(8C)(5C,)(8C;) |(5x 56)+ (10x 70)+(10x 56) + (5x 28)+(1x 8) + 1 - 280700560 140+8 =1,688

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