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Question Number 160980 by cortano last updated on 10/Dec/21

 How many ways can 50 people   be divided into 3 groups, so   that each group contains members   equal to a prime number?

$$\:\mathrm{How}\:\mathrm{many}\:\mathrm{ways}\:\mathrm{can}\:\mathrm{50}\:\mathrm{people} \\ $$$$\:\mathrm{be}\:\mathrm{divided}\:\mathrm{into}\:\mathrm{3}\:\mathrm{groups},\:\mathrm{so} \\ $$$$\:\mathrm{that}\:\mathrm{each}\:\mathrm{group}\:\mathrm{contains}\:\mathrm{members} \\ $$$$\:\mathrm{equal}\:\mathrm{to}\:\mathrm{a}\:\mathrm{prime}\:\mathrm{number}? \\ $$

Answered by Rasheed.Sindhi last updated on 10/Dec/21

Sum of three primes=50  ⇒One is even prime and the remaining  two are odd primes  So one group is of 2-people  The remaining 48 should be divided  into two ′prime groups.  48=7+41,11+37,17+31,19+29  Hence 4 ways.  (2,7,41),(2,11,37),(2,17,31),(2,19,29)

$$\mathrm{Sum}\:\mathrm{of}\:\mathrm{three}\:\mathrm{primes}=\mathrm{50} \\ $$$$\Rightarrow\mathrm{One}\:\mathrm{is}\:\mathrm{even}\:\mathrm{prime}\:\mathrm{and}\:\mathrm{the}\:\mathrm{remaining} \\ $$$$\mathrm{two}\:\mathrm{are}\:\mathrm{odd}\:\mathrm{primes} \\ $$$$\mathrm{So}\:\mathrm{one}\:\mathrm{group}\:\mathrm{is}\:\mathrm{of}\:\mathrm{2}-\mathrm{people} \\ $$$$\mathrm{The}\:\mathrm{remaining}\:\mathrm{48}\:\mathrm{should}\:\mathrm{be}\:\mathrm{divided} \\ $$$$\mathrm{into}\:\mathrm{two}\:'\mathrm{prime}\:\mathrm{groups}. \\ $$$$\mathrm{48}=\mathrm{7}+\mathrm{41},\mathrm{11}+\mathrm{37},\mathrm{17}+\mathrm{31},\mathrm{19}+\mathrm{29} \\ $$$${Hence}\:\mathrm{4}\:{ways}. \\ $$$$\left(\mathrm{2},\mathrm{7},\mathrm{41}\right),\left(\mathrm{2},\mathrm{11},\mathrm{37}\right),\left(\mathrm{2},\mathrm{17},\mathrm{31}\right),\left(\mathrm{2},\mathrm{19},\mathrm{29}\right) \\ $$

Commented by cortano last updated on 10/Dec/21

not 30 ways ?

$$\mathrm{not}\:\mathrm{30}\:\mathrm{ways}\:? \\ $$

Commented by Rasheed.Sindhi last updated on 10/Dec/21

I think all the permutations of(2,7,41)  one way.

$${I}\:{think}\:{all}\:{the}\:{permutations}\:{of}\left(\mathrm{2},\mathrm{7},\mathrm{41}\right) \\ $$$${one}\:{way}.\: \\ $$

Commented by bobhans last updated on 10/Dec/21

how for (2,5,43)?

$$\mathrm{how}\:\mathrm{for}\:\left(\mathrm{2},\mathrm{5},\mathrm{43}\right)? \\ $$

Commented by Rasheed.Sindhi last updated on 10/Dec/21

Yes this also. Thank you!

$${Yes}\:{this}\:{also}.\:{Thank}\:{you}! \\ $$

Answered by mr W last updated on 10/Dec/21

Commented by mr W last updated on 10/Dec/21

a+b+c=50 with a,b,c ∈P  except 2 all prime numbers are odd.  the sum of three odd numbers is   always odd. such that the  sum of three prime numbers is 50,   one and only one of them must be 2.  say a=2, then b+c=48. from the  prime number table above we see  there are following possibilities:  43+5  41+7  37+11  31+17  29+19    to divide 50 people in three groups  with a, b, c people in each group  respectively there are ((50!)/(a!b!c!)) ways.  so the total number of ways is:  ((50!)/(2!5!43!))+((50!)/(2!7!41!))+((50!)/(2!11!37!))+((50!)/(2!17!31!))+((50!)/(2!19!29!))  ≈1.94×10^(16)

$${a}+{b}+{c}=\mathrm{50}\:{with}\:{a},{b},{c}\:\in\mathbb{P} \\ $$$${except}\:\mathrm{2}\:{all}\:{prime}\:{numbers}\:{are}\:{odd}. \\ $$$${the}\:{sum}\:{of}\:{three}\:{odd}\:{numbers}\:{is}\: \\ $$$${always}\:{odd}.\:{such}\:{that}\:{the} \\ $$$${sum}\:{of}\:{three}\:{prime}\:{numbers}\:{is}\:\mathrm{50},\: \\ $$$${one}\:{and}\:{only}\:{one}\:{of}\:{them}\:{must}\:{be}\:\mathrm{2}. \\ $$$${say}\:{a}=\mathrm{2},\:{then}\:{b}+{c}=\mathrm{48}.\:{from}\:{the} \\ $$$${prime}\:{number}\:{table}\:{above}\:{we}\:{see} \\ $$$${there}\:{are}\:{following}\:{possibilities}: \\ $$$$\mathrm{43}+\mathrm{5} \\ $$$$\mathrm{41}+\mathrm{7} \\ $$$$\mathrm{37}+\mathrm{11} \\ $$$$\mathrm{31}+\mathrm{17} \\ $$$$\mathrm{29}+\mathrm{19} \\ $$$$ \\ $$$${to}\:{divide}\:\mathrm{50}\:{people}\:{in}\:{three}\:{groups} \\ $$$${with}\:{a},\:{b},\:{c}\:{people}\:{in}\:{each}\:{group} \\ $$$${respectively}\:{there}\:{are}\:\frac{\mathrm{50}!}{{a}!{b}!{c}!}\:{ways}. \\ $$$${so}\:{the}\:{total}\:{number}\:{of}\:{ways}\:{is}: \\ $$$$\frac{\mathrm{50}!}{\mathrm{2}!\mathrm{5}!\mathrm{43}!}+\frac{\mathrm{50}!}{\mathrm{2}!\mathrm{7}!\mathrm{41}!}+\frac{\mathrm{50}!}{\mathrm{2}!\mathrm{11}!\mathrm{37}!}+\frac{\mathrm{50}!}{\mathrm{2}!\mathrm{17}!\mathrm{31}!}+\frac{\mathrm{50}!}{\mathrm{2}!\mathrm{19}!\mathrm{29}!} \\ $$$$\approx\mathrm{1}.\mathrm{94}×\mathrm{10}^{\mathrm{16}} \\ $$

Commented by Tawa11 last updated on 10/Dec/21

Great sir

$$\mathrm{Great}\:\mathrm{sir} \\ $$

Commented by cortano last updated on 10/Dec/21

thank you

$$\mathrm{thank}\:\mathrm{you} \\ $$

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