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Question Number 11384 by tawa last updated on 22/Mar/17

Out of 5 accountants and 7 bankers, a committee consisting of   2 accountants and 3 bankers is to be formed. In how many ways can this  be done if  (a) Any acountant and any bankers must be included  (b) One particular banker must be included  (c) 2 accountant cannot be in the committee

$$\mathrm{Out}\:\mathrm{of}\:\mathrm{5}\:\mathrm{accountants}\:\mathrm{and}\:\mathrm{7}\:\mathrm{bankers},\:\mathrm{a}\:\mathrm{committee}\:\mathrm{consisting}\:\mathrm{of}\: \\ $$$$\mathrm{2}\:\mathrm{accountants}\:\mathrm{and}\:\mathrm{3}\:\mathrm{bankers}\:\mathrm{is}\:\mathrm{to}\:\mathrm{be}\:\mathrm{formed}.\:\mathrm{In}\:\mathrm{how}\:\mathrm{many}\:\mathrm{ways}\:\mathrm{can}\:\mathrm{this} \\ $$$$\mathrm{be}\:\mathrm{done}\:\mathrm{if} \\ $$$$\left(\mathrm{a}\right)\:\mathrm{Any}\:\mathrm{acountant}\:\mathrm{and}\:\mathrm{any}\:\mathrm{bankers}\:\mathrm{must}\:\mathrm{be}\:\mathrm{included} \\ $$$$\left(\mathrm{b}\right)\:\mathrm{One}\:\mathrm{particular}\:\mathrm{banker}\:\mathrm{must}\:\mathrm{be}\:\mathrm{included} \\ $$$$\left(\mathrm{c}\right)\:\mathrm{2}\:\mathrm{accountant}\:\mathrm{cannot}\:\mathrm{be}\:\mathrm{in}\:\mathrm{the}\:\mathrm{committee} \\ $$

Answered by sandy_suhendra last updated on 23/Mar/17

(a) 4C1 × 6C2 =4×15 = 60  (b) 5C2 × 6C2 = 10×15 = 150  (c) 3C2 × 7C3 = 3 × 35 = 105

$$\left(\mathrm{a}\right)\:\mathrm{4C1}\:×\:\mathrm{6C2}\:=\mathrm{4}×\mathrm{15}\:=\:\mathrm{60} \\ $$$$\left(\mathrm{b}\right)\:\mathrm{5C2}\:×\:\mathrm{6C2}\:=\:\mathrm{10}×\mathrm{15}\:=\:\mathrm{150} \\ $$$$\left(\mathrm{c}\right)\:\mathrm{3C2}\:×\:\mathrm{7C3}\:=\:\mathrm{3}\:×\:\mathrm{35}\:=\:\mathrm{105} \\ $$

Commented by tawa last updated on 23/Mar/17

God bless you sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

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