Question Number 139741 by mathocean1 last updated on 30/Apr/21
$${A}\:{man}\:{invite}\:\mathrm{5}\:{friends}\:{choosen} \\ $$$${from}\:\mathrm{10}\:{for}\:{a}\:{diner}.\:{He}\:{don}'{t}\: \\ $$$${want}\:{that}\:{two}\:{of}\:{them}\:{took}\:{part} \\ $$$${to}\:{that}\:{diner}.\:{The}\:{number}\:{of} \\ $$$${ways}\:{to}\:{choose}\:{her}\:{friends}\:{is}: \\ $$$$\left.{a}\right)\mathrm{296} \\ $$$$\left.{b}\right)\mathrm{196} \\ $$$$\left.{c}\right)\mathrm{5} \\ $$$$\left.{d}\right)\mathrm{252} \\ $$
Answered by mr W last updated on 01/May/21
$${A}\:{is}\:{selected}\:{but}\:{B}\:{not}:\:{C}_{\mathrm{4}} ^{\mathrm{8}} \:{ways} \\ $$$${B}\:{is}\:{selected}\:{but}\:{A}\:{not}:\:{C}_{\mathrm{4}} ^{\mathrm{8}} \:{ways} \\ $$$${both}\:{A}\:{and}\:{B}\:{are}\:{not}\:{selected}:\:{C}_{\mathrm{5}} ^{\mathrm{8}} \:{ways} \\ $$$${totally}:\:\mathrm{2}×{C}_{\mathrm{4}} ^{\mathrm{8}} +{C}_{\mathrm{5}} ^{\mathrm{8}} =\mathrm{196}\:{ways}. \\ $$$$ \\ $$$${or} \\ $$$${to}\:{select}\:\mathrm{5}\:{from}\:\mathrm{10}:\:{C}_{\mathrm{5}} ^{\mathrm{10}} \:{ways} \\ $$$${minus} \\ $$$${both}\:{A}\:{and}\:{B}\:{are}\:{selected}:\:{C}_{\mathrm{3}} ^{\mathrm{8}} \\ $$$${valid}:\:{C}_{\mathrm{5}} ^{\mathrm{10}} −{C}_{\mathrm{3}} ^{\mathrm{8}} =\mathrm{196}\:{ways} \\ $$
Commented by otchereabdullai@gmail.com last updated on 01/May/21
$$\mathrm{nice}\:\mathrm{one}! \\ $$
Answered by mathocean1 last updated on 01/May/21
$${thanks} \\ $$