The number of ways in which 30 coins of one rupee each be given to six persons so that none them receive less than 4 rupees is:

Options

A. 231
B. 462
C. 493
D. 924

Answer

To solve the problem of distributing 30 identical coins of one rupee each to six persons such that none of them receives less than 4 rupees, we can follow these steps:

### Step 1: Understand the Problem
We have 30 coins and 6 persons. Each person must receive at least 4 coins.

### Step 2: Allocate Minimum Coins
Since each of the 6 persons must receive at least 4 coins, we can first allocate 4 coins to each person. This means we distribute:
\[ 4 \times 6 = 24 \text{ coins} \]
Now, we have:
\[ 30 - 24 = 6 \text{ coins remaining} \]

### Step 3: Reformulate the Problem
Now, we need to distribute the remaining 6 coins among the 6 persons with no restrictions (i.e., a person can receive 0 or more coins). Let \( x_1, x_2, x_3, x_4, x_5, x_6 \) represent the number of additional coins received by each person. We need to solve the equation:
\[ x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 6 \]
where \( x_i \geq 0 \) for each \( i \).

### Step 4: Apply the Stars and Bars Theorem
The problem can be solved using the "stars and bars" theorem, which states that the number of ways to distribute \( n \) identical items (stars) into \( r \) distinct groups (bars) is given by:
\[ \binom{n + r - 1}{r - 1} \]
In our case, \( n = 6 \) (the remaining coins) and \( r = 6 \) (the persons). Thus, we need to calculate:
\[ \binom{6 + 6 - 1}{6 - 1} = \binom{11}{5} \]

### Step 5: Calculate \( \binom{11}{5} \)
Now we calculate \( \binom{11}{5} \):
\[ \binom{11}{5} = \frac{11!}{5! \cdot (11 - 5)!} = \frac{11!}{5! \cdot 6!} \]
Calculating the factorials:
\[ = \frac{11 \times 10 \times 9 \times 8 \times 7}{5 \times 4 \times 3 \times 2 \times 1} = \frac{55440}{120} = 462 \]

### Final Answer
Thus, the number of ways in which 30 coins can be distributed to 6 persons such that none of them receives less than 4 rupees is:
\[ \boxed{462} \]