Question

The number of seven-digit positive integers formed using the digits 1, 2, 3, and 4 only, and whose sum of the digits is 12, is         

• A. 413

Answer 1

The Correct Option is A

We need to find the number of seven-digit positive integers formed using the digits 1, 2, 3, and 4, such that the sum of the digits is equal to 12.

Let the digits of the seven-digit integer be \( x_1, x_2, x_3, x_4, x_5, x_6, x_7 \), where each \( x_i \in \{1, 2, 3, 4\} \). The equation we are solving is: \[ x_1 + x_2 + x_3 + x_4 + x_5 + x_6 + x_7 = 12. \]

We want to count the number of solutions to this equation, subject to the condition \( 1 \leq x_i \leq 4 \) for each \( i \).

This is a problem of counting the number of solutions to a Diophantine equation, and it can be solved using the stars and bars method with adjustments for the restrictions on the values of \( x_i \).

Let \( y_i = x_i - 1 \). Then \( x_i = y_i + 1 \), and \( 0 \leq y_i \leq 3 \). Substitute this into the equation: \[ (y_1 + 1) + (y_2 + 1) + \cdots + (y_7 + 1) = 12 \] \[ y_1 + y_2 + \cdots + y_7 = 12 - 7 = 5 \]

Now we need to find the number of non-negative integer solutions to \( y_1 + y_2 + \cdots + y_7 = 5 \). Using stars and bars, the number of solutions is: \[ \binom{5 + 7 - 1}{7 - 1} = \binom{11}{6} = 462 \]

However, we have the restriction \( 0 \leq y_i \leq 3 \). We need to subtract the cases where any \( y_i \geq 4 \). Since the total sum is 5, at most one variable can be greater than or equal to 4.

Cases where one variable is greater than or equal to 4: Let \( y_1 \geq 4 \). Let \( z_1 = y_1 - 4 \). Then \( z_1 + y_2 + \cdots + y_7 = 1 \). The number of solutions is \(\binom{1 + 7 - 1}{7 - 1} = \binom{7}{6} = 7\). Since any of the 7 variables could be greater than or equal to 4, we multiply by 7: \( 7 \times 7 = 49 \).

Subtract the invalid cases from the initial solution: \[ 462 - 49 = 413 \]

Therefore, the correct answer is 413.

There are 413 such seven-digit integers.