A committee of 3 experts is to be selected out of a panel of 7 persons. Three of them are engineers, Three are managers and, one is both engineer and manager. In how many ways can the committee be selected if it must have atleast one engineer and one manager?
Number of ways the committee be selected with atleast one engineer and one manager,
= $$3_{C_{1}}3_{C_{1}}3_{C_{1}} +Â 3_{C_{1}}3_{C_{2}} + 3_{C_{2}}3_{C_{1}} +Â 3_{C_{2}}1_{C_{1}} + 3_{C_{2}}1_{C_{1}}$$
= 9 + 9 + 9 + 3 + 3 = 33
Hence, option A is the correct answer.
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