Question Number 160980 by cortano last updated on 10/Dec/21
$$\:\mathrm{How}\:\mathrm{many}\:\mathrm{ways}\:\mathrm{can}\:\mathrm{50}\:\mathrm{people} \\ $$$$\:\mathrm{be}\:\mathrm{divided}\:\mathrm{into}\:\mathrm{3}\:\mathrm{groups},\:\mathrm{so} \\ $$$$\:\mathrm{that}\:\mathrm{each}\:\mathrm{group}\:\mathrm{contains}\:\mathrm{members} \\ $$$$\:\mathrm{equal}\:\mathrm{to}\:\mathrm{a}\:\mathrm{prime}\:\mathrm{number}? \\ $$
Answered by Rasheed.Sindhi last updated on 10/Dec/21
$$\mathrm{Sum}\:\mathrm{of}\:\mathrm{three}\:\mathrm{primes}=\mathrm{50} \\ $$$$\Rightarrow\mathrm{One}\:\mathrm{is}\:\mathrm{even}\:\mathrm{prime}\:\mathrm{and}\:\mathrm{the}\:\mathrm{remaining} \\ $$$$\mathrm{two}\:\mathrm{are}\:\mathrm{odd}\:\mathrm{primes} \\ $$$$\mathrm{So}\:\mathrm{one}\:\mathrm{group}\:\mathrm{is}\:\mathrm{of}\:\mathrm{2}-\mathrm{people} \\ $$$$\mathrm{The}\:\mathrm{remaining}\:\mathrm{48}\:\mathrm{should}\:\mathrm{be}\:\mathrm{divided} \\ $$$$\mathrm{into}\:\mathrm{two}\:'\mathrm{prime}\:\mathrm{groups}. \\ $$$$\mathrm{48}=\mathrm{7}+\mathrm{41},\mathrm{11}+\mathrm{37},\mathrm{17}+\mathrm{31},\mathrm{19}+\mathrm{29} \\ $$$${Hence}\:\mathrm{4}\:{ways}. \\ $$$$\left(\mathrm{2},\mathrm{7},\mathrm{41}\right),\left(\mathrm{2},\mathrm{11},\mathrm{37}\right),\left(\mathrm{2},\mathrm{17},\mathrm{31}\right),\left(\mathrm{2},\mathrm{19},\mathrm{29}\right) \\ $$
Commented by cortano last updated on 10/Dec/21
$$\mathrm{not}\:\mathrm{30}\:\mathrm{ways}\:? \\ $$
Commented by Rasheed.Sindhi last updated on 10/Dec/21
$${I}\:{think}\:{all}\:{the}\:{permutations}\:{of}\left(\mathrm{2},\mathrm{7},\mathrm{41}\right) \\ $$$${one}\:{way}.\: \\ $$
Commented by bobhans last updated on 10/Dec/21
$$\mathrm{how}\:\mathrm{for}\:\left(\mathrm{2},\mathrm{5},\mathrm{43}\right)? \\ $$
Commented by Rasheed.Sindhi last updated on 10/Dec/21
$${Yes}\:{this}\:{also}.\:{Thank}\:{you}! \\ $$
Answered by mr W last updated on 10/Dec/21
Commented by mr W last updated on 10/Dec/21
$${a}+{b}+{c}=\mathrm{50}\:{with}\:{a},{b},{c}\:\in\mathbb{P} \\ $$$${except}\:\mathrm{2}\:{all}\:{prime}\:{numbers}\:{are}\:{odd}. \\ $$$${the}\:{sum}\:{of}\:{three}\:{odd}\:{numbers}\:{is}\: \\ $$$${always}\:{odd}.\:{such}\:{that}\:{the} \\ $$$${sum}\:{of}\:{three}\:{prime}\:{numbers}\:{is}\:\mathrm{50},\: \\ $$$${one}\:{and}\:{only}\:{one}\:{of}\:{them}\:{must}\:{be}\:\mathrm{2}. \\ $$$${say}\:{a}=\mathrm{2},\:{then}\:{b}+{c}=\mathrm{48}.\:{from}\:{the} \\ $$$${prime}\:{number}\:{table}\:{above}\:{we}\:{see} \\ $$$${there}\:{are}\:{following}\:{possibilities}: \\ $$$$\mathrm{43}+\mathrm{5} \\ $$$$\mathrm{41}+\mathrm{7} \\ $$$$\mathrm{37}+\mathrm{11} \\ $$$$\mathrm{31}+\mathrm{17} \\ $$$$\mathrm{29}+\mathrm{19} \\ $$$$ \\ $$$${to}\:{divide}\:\mathrm{50}\:{people}\:{in}\:{three}\:{groups} \\ $$$${with}\:{a},\:{b},\:{c}\:{people}\:{in}\:{each}\:{group} \\ $$$${respectively}\:{there}\:{are}\:\frac{\mathrm{50}!}{{a}!{b}!{c}!}\:{ways}. \\ $$$${so}\:{the}\:{total}\:{number}\:{of}\:{ways}\:{is}: \\ $$$$\frac{\mathrm{50}!}{\mathrm{2}!\mathrm{5}!\mathrm{43}!}+\frac{\mathrm{50}!}{\mathrm{2}!\mathrm{7}!\mathrm{41}!}+\frac{\mathrm{50}!}{\mathrm{2}!\mathrm{11}!\mathrm{37}!}+\frac{\mathrm{50}!}{\mathrm{2}!\mathrm{17}!\mathrm{31}!}+\frac{\mathrm{50}!}{\mathrm{2}!\mathrm{19}!\mathrm{29}!} \\ $$$$\approx\mathrm{1}.\mathrm{94}×\mathrm{10}^{\mathrm{16}} \\ $$
Commented by Tawa11 last updated on 10/Dec/21
$$\mathrm{Great}\:\mathrm{sir} \\ $$
Commented by cortano last updated on 10/Dec/21
$$\mathrm{thank}\:\mathrm{you} \\ $$