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How many distinct ways are there to split $50$ identical coins among three people so that each person gets at least $5$ coins?

  1. $3^{35}$
  2. $3^{50}-2^{50}$
  3. $\binom{35}{2}$
  4. $\binom{50}{15} \cdot 3^{35}$
  5. $\binom{37}{2}$
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4 Answers

Best answer
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32 votes

Distinct ways are there to split $50$ identical coins among three people so that each person gets at least $5$ coins

$x_{1}+5+x_{2}+5+x_{3}+5= 50$

$x_{1}+x_{2}+x_{3} = 35$

Solving Non-integral solution $n=35 ,r =3$

$^{\left(n+r-1\right)}C_{r-1}= ^{\left(35+3-1\right)}C_{3-1} = ^{37}C_{2}.$

Hence E is Answer

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$1^{st}$  $5$ coins are distributed to everyone .... so  $50-15= 35$ coins remain.

Now these $35$ coins will be distributed among $3$ people arbitrarily ...

$\therefore\ x_1+x_2+x_3= 35$

Using Generating function it will be coefficient of $x^{35}$

$[x^{35}] [x^0+x^1+....+x^{35}]^3$  [$\because$ $35$ coins are distributed among $3$ people]

$=[x^{35}] [1-x^{36}]^3 [1-x]^{-3}$

solving it The coefficient will be $^{3+35-1}C_{35} =\ ^{37}C_2$

Hence $E$ is the ans .. Correct me if i am wrong ...
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This is equivalent to say Distribution of  50 non-distinct coins into 3 distinct people where each people gets atleast 5 coins.

First, distribute 5 coins to each people.Then the number of coins left is 35.Then Distribute 35 coins to 3 peoples arbitrarily.

So, total no of ways of Distributing 35 coins to 3 peoples arbitrarily (i.e people get 0 or more coins) = C(3+35-1, 35)  = C(37, 35) =  C(37, 2) = 666

The correct answer is,(E) C(37, 2)

Answer:

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