a. equal number of each
Combinations again..
architects 4!/(2!2!)
engineers 6!/(2!4!)
electricians 9!/(2!7!)
b. remove the 2 engineers from the choices which would leave 4 architechts, 4 engineers and 9 electricians Now, we have 17 to choose from to make a group of 6.
17!/(6! 11!)
Out of 4 architects, 6 engineers and 9 electricians, how many ways can a group of 6 people be formed to attend a forum if:
a. there is an equal number of architects, engineers and electricians
b. two particular engineers cannot be in the committee?
c. there are at least 3 electricians?
2 answers
Thanks much :)