Answer
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Hint:
We can distribute 3 coins each to the 5 persons. Then we can find the number of remaining coins by subtracting it from the total coins. Then we can find the number of ways of distributing the remaining coins to 5 people using the formula ${}^{n + r - 1}{C_{r - 1}}$ where n things are distributed to r things. Then this number of ways will give the required number of ways.
Complete step by step solution:
We have 20 coins and we need to distribute it to 5 people. It is given that each person must get at least 3 rupees. As all the coins are one-rupee coins, we can say that each person must have 3 coins each.
Now we can distribute 3 coins to each person out of the 5 people. Then the number of remaining coins is given by.
$ \Rightarrow n = 20 - 3 \times 5$
On simplification, we get
$ \Rightarrow n = 20 - 15$
So, we have
$ \Rightarrow n = 5$
Now, we need to distribute the remaining 5 coins to 5 people without any conditions. So, the number of ways of distributing the remaining 5 coins to 5 people will give the required number of ways of distributing 20 coins among 5 people such that each person gets at least 3 rupees.
We know that n objects can be distributed to r objects in ${}^{n + r - 1}{C_{r - 1}}$ . Here $r = 5$ and $n = 5$ . So, the required number of ways is given by,
$N = {}^{5 + 5 - 1}{C_{5 - 1}}$
On simplification, we get
$ \Rightarrow N = {}^9{C_4}$
We know that the expansion of the combinations is given by, ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ .
$ \Rightarrow N = \dfrac{{9!}}{{4!\left( {9 - 4} \right)!}}$
On simplification, we get
$ \Rightarrow N = \dfrac{{9!}}{{4! \times 5!}}$
Now, we can expand the factorial
$ \Rightarrow N = \dfrac{{9 \times 8 \times 7 \times 6 \times 5!}}{{4 \times 3 \times 2 \times 5!}}$
On cancelling the common terms, we get
$ \Rightarrow N = 9 \times 2 \times 7$
On multiplication, we have
$ \Rightarrow N = 126$
Therefore, the required number of ways of distributing 20 coins among 5 people such that each person gets at least 3 rupees is 126.
So, the correct answer is option A.
Note:
We cannot find the number of ways directly by taking different cases as there will be a lot of cases and we can make errors. We must note that the equation used for finding the number of ways of distribution is different from the regular combinations and permutations which involves selecting. After distributing 3 coins to each person, we must only consider the rest of the coins and not the total number of coins.
We can distribute 3 coins each to the 5 persons. Then we can find the number of remaining coins by subtracting it from the total coins. Then we can find the number of ways of distributing the remaining coins to 5 people using the formula ${}^{n + r - 1}{C_{r - 1}}$ where n things are distributed to r things. Then this number of ways will give the required number of ways.
Complete step by step solution:
We have 20 coins and we need to distribute it to 5 people. It is given that each person must get at least 3 rupees. As all the coins are one-rupee coins, we can say that each person must have 3 coins each.
Now we can distribute 3 coins to each person out of the 5 people. Then the number of remaining coins is given by.
$ \Rightarrow n = 20 - 3 \times 5$
On simplification, we get
$ \Rightarrow n = 20 - 15$
So, we have
$ \Rightarrow n = 5$
Now, we need to distribute the remaining 5 coins to 5 people without any conditions. So, the number of ways of distributing the remaining 5 coins to 5 people will give the required number of ways of distributing 20 coins among 5 people such that each person gets at least 3 rupees.
We know that n objects can be distributed to r objects in ${}^{n + r - 1}{C_{r - 1}}$ . Here $r = 5$ and $n = 5$ . So, the required number of ways is given by,
$N = {}^{5 + 5 - 1}{C_{5 - 1}}$
On simplification, we get
$ \Rightarrow N = {}^9{C_4}$
We know that the expansion of the combinations is given by, ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ .
$ \Rightarrow N = \dfrac{{9!}}{{4!\left( {9 - 4} \right)!}}$
On simplification, we get
$ \Rightarrow N = \dfrac{{9!}}{{4! \times 5!}}$
Now, we can expand the factorial
$ \Rightarrow N = \dfrac{{9 \times 8 \times 7 \times 6 \times 5!}}{{4 \times 3 \times 2 \times 5!}}$
On cancelling the common terms, we get
$ \Rightarrow N = 9 \times 2 \times 7$
On multiplication, we have
$ \Rightarrow N = 126$
Therefore, the required number of ways of distributing 20 coins among 5 people such that each person gets at least 3 rupees is 126.
So, the correct answer is option A.
Note:
We cannot find the number of ways directly by taking different cases as there will be a lot of cases and we can make errors. We must note that the equation used for finding the number of ways of distribution is different from the regular combinations and permutations which involves selecting. After distributing 3 coins to each person, we must only consider the rest of the coins and not the total number of coins.
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