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Question Number 138748 by physicstutes last updated on 17/Apr/21

A committee of 3 members is to be formed from 8 members. Find the number of committees that can be formed if two particular club members cannot both be in a committee

$$\mathrm{A}\:\mathrm{committee}\:\mathrm{of}\:\mathrm{3}\:\mathrm{members}\:\mathrm{is}\:\mathrm{to}\:\mathrm{be}\:\mathrm{formed}\:\mathrm{from}\:\mathrm{8}\:\mathrm{members}. \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{committees}\:\mathrm{that}\:\mathrm{can}\:\mathrm{be}\:\mathrm{formed}\:\mathrm{if}\:\mathrm{two}\:\mathrm{particular} \\ $$$$\mathrm{club}\:\mathrm{members}\:\mathrm{cannot}\:\mathrm{both}\:\mathrm{be}\:\mathrm{in}\:\mathrm{a}\:\mathrm{committee} \\ $$

Answered by mr W last updated on 17/Apr/21

C_3 ^8 −C_1 ^6 =50

$${C}_{\mathrm{3}} ^{\mathrm{8}} −{C}_{\mathrm{1}} ^{\mathrm{6}} =\mathrm{50} \\ $$

Commented by physicstutes last updated on 17/Apr/21

sir please why the ^6 C_1 part?

$$\mathrm{sir}\:\mathrm{please}\:\mathrm{why}\:\mathrm{the}\:\:^{\mathrm{6}} \mathrm{C}_{\mathrm{1}} \:\mathrm{part}? \\ $$

Commented by mr W last updated on 17/Apr/21

to select 3 from 8 persons there are totally C_3 ^8 ways. if two particular persons are in the committee, to select the third person from the rest 6 persons, there are C_1 ^6 ways. since these two particular persons should not be both in the committee, so we have only C_3 ^8 −C_1 ^6 ways.

$${to}\:{select}\:\mathrm{3}\:{from}\:\mathrm{8}\:{persons}\:{there}\:{are} \\ $$$${totally}\:{C}_{\mathrm{3}} ^{\mathrm{8}} \:{ways}. \\ $$$${if}\:{two}\:{particular}\:{persons}\:{are}\:{in}\:{the} \\ $$$${committee},\:{to}\:{select}\:{the}\:{third}\:{person} \\ $$$${from}\:{the}\:{rest}\:\mathrm{6}\:{persons},\:{there}\:{are} \\ $$$${C}_{\mathrm{1}} ^{\mathrm{6}} \:{ways}. \\ $$$${since}\:{these}\:{two}\:{particular}\:{persons} \\ $$$${should}\:{not}\:{be}\:{both}\:{in}\:{the}\:{committee}, \\ $$$${so}\:{we}\:{have}\:{only}\:{C}_{\mathrm{3}} ^{\mathrm{8}} −{C}_{\mathrm{1}} ^{\mathrm{6}} \:{ways}. \\ $$

Commented by mr W last updated on 17/Apr/21

certainly you can also solve like this: say the two particular persons are A and B. 1) both A and B are not in committee we select all three members from the other 6 persons, there are C_3 ^6 ways. 2) A is in committe we select other two members from the other 6 persons, there are C_2 ^6 ways. 3) B is in committe we select other two members from the other 6 persons, there are C_2 ^6 ways. totally: there are C_3 ^6 +2×C_2 ^6 =50 ways.

$${certainly}\:{you}\:{can}\:{also}\:{solve}\:{like}\:{this}: \\ $$$${say}\:{the}\:{two}\:{particular}\:{persons}\:{are} \\ $$$${A}\:{and}\:{B}. \\ $$$$\left.\mathrm{1}\right)\:{both}\:{A}\:{and}\:{B}\:{are}\:{not}\:{in}\:{committee} \\ $$$${we}\:{select}\:{all}\:{three}\:{members}\:{from}\:{the} \\ $$$${other}\:\mathrm{6}\:{persons},\:{there}\:{are}\:{C}_{\mathrm{3}} ^{\mathrm{6}} \:{ways}. \\ $$$$\left.\mathrm{2}\right)\:{A}\:{is}\:{in}\:{committe} \\ $$$${we}\:{select}\:{other}\:{two}\:{members}\:{from} \\ $$$${the}\:{other}\:\mathrm{6}\:{persons},\:{there}\:{are}\:{C}_{\mathrm{2}} ^{\mathrm{6}} \:{ways}. \\ $$$$\left.\mathrm{3}\right)\:{B}\:{is}\:{in}\:{committe} \\ $$$${we}\:{select}\:{other}\:{two}\:{members}\:{from} \\ $$$${the}\:{other}\:\mathrm{6}\:{persons},\:{there}\:{are}\:{C}_{\mathrm{2}} ^{\mathrm{6}} \:{ways}. \\ $$$${totally}:\:{there}\:{are} \\ $$$${C}_{\mathrm{3}} ^{\mathrm{6}} +\mathrm{2}×{C}_{\mathrm{2}} ^{\mathrm{6}} =\mathrm{50}\:{ways}. \\ $$

Commented by physicstutes last updated on 18/Apr/21

that was brilliant sir. Thanks

$$\mathrm{that}\:\mathrm{was}\:\mathrm{brilliant}\:\mathrm{sir}.\:\mathrm{Thanks} \\ $$

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