Question Number 56961 by pete last updated on 27/Mar/19
$$\mathrm{Out}\:\mathrm{of}\:\mathrm{6}\:\mathrm{mathematicians}\:\mathrm{and}\:\mathrm{7}\:\mathrm{physicists} \\ $$$$\mathrm{a}\:\mathrm{committee}\:\mathrm{consisting}\:\mathrm{of}\:\mathrm{3}\:\mathrm{mathematicians} \\ $$$$\mathrm{and}\:\mathrm{3}\:\mathrm{physicists}\:\mathrm{is}\:\mathrm{to}\:\mathrm{be}\:\mathrm{formed}.\:\mathrm{In}\:\mathrm{how} \\ $$$$\mathrm{many}\:\mathrm{ways}\:\mathrm{can}\:\mathrm{this}\:\mathrm{be}\:\mathrm{done}\:\mathrm{if}\:\mathrm{two}\: \\ $$$$\mathrm{particular}\:\mathrm{mathematicians}\:\mathrm{cannot}\:\mathrm{be} \\ $$$$\mathrm{on}\:\mathrm{the}\:\mathrm{commitee}? \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 27/Mar/19
$${let}\:{mr}.{A}_{{math}} \:\:{and}\:{mr}.{B}_{{math}} \:{can}\:{not}\:{be}\:{on}\:{the} \\ $$$${commitee}. \\ $$$$ \\ $$$${if}\:{mr}.{A}\:{and}\:{mr}.{B}\:{are}\:{not}\:{in}\:{commitee}\:{at}\:{all} \\ $$$${then}\:{available}\:{mathematician}\:\left(\mathrm{6}−\mathrm{2}=\mathrm{4}\right) \\ $$$${answer}\:{is}\:\mathrm{4}{C}_{\mathrm{3}} ×\mathrm{7}{C}_{\mathrm{3}} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by pete last updated on 27/Mar/19
$$\mathrm{Thanks}\:\mathrm{Sir} \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 27/Mar/19
$${most}\:{welcome}… \\ $$