JEE Mathematics 2023 Question with Solution

Given: 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 $12$.

Find: The total number of such integers.

Let the digits be $x_1, x_2, x_3, x_4, x_5, x_6, x_7$, where each $x_i \in \{1,2,3,4\}$.

Then

$$x_1 + x_2 + x_3 + x_4 + x_5 + x_6 + x_7 = 12$$

We want the number of solutions subject to $1 \le x_i \le 4$ for each $i$.

Let

$$y_i = x_i - 1$$

Then $x_i = y_i + 1$ and $0 \le y_i \le 3$.

Substituting,

$$(y_1 + 1) + (y_2 + 1) + \cdots + (y_7 + 1) = 12$$

so

$$y_1 + y_2 + \cdots + y_7 = 5$$

First count all non-negative integer solutions:

$$\binom{5+7-1}{7-1} = \binom{11}{6} = 462$$

Now subtract the cases where some $y_i \ge 4$.

Since the total sum is only $5$, at most one variable can be at least $4$.

If one variable, say $y_1 \ge 4$, let

$$z_1 = y_1 - 4$$

Then

$$z_1 + y_2 + \cdots + y_7 = 1$$

The number of solutions is

$$\binom{1+7-1}{7-1} = \binom{7}{6} = 7$$

Any one of the $7$ variables could be the variable exceeding or equal to $4$, so the number of invalid cases is

$$7 \times 7 = 49$$

Hence the required count is

$$462 - 49 = 413$$

Therefore, the number of such seven-digit integers is $413$.

Given: A seven-digit number uses only the digits $1,2,3,4$ and has digit sum $12$.

Find: The number of such numbers.

Because every digit is at least $1$, start by assigning $1$ to each of the $7$ positions. This contributes

$$7$$

to the digit sum.

So we still need an additional

$$12 - 7 = 5$$

to be distributed among the $7$ positions.

If the extra amount added to the $i$-th digit is $y_i$, then

$$y_1 + y_2 + \cdots + y_7 = 5$$

with

$$0 \le y_i \le 3$$

because the original digit cannot exceed $4$.

Ignoring the upper bound for a moment, the number of non-negative integer solutions is

$$\binom{11}{6} = 462$$

Now remove forbidden cases.

A forbidden case occurs when some $y_i \ge 4$. Since the total sum is only $5$, two variables cannot both be at least $4$. So only one-variable violations are possible.

Choose the violating variable in

$$7$$

ways.

After forcing that variable to contribute $4$, the remaining sum is

$$5 - 4 = 1$$

which can be distributed among $7$ variables in

$$\binom{7}{6} = 7$$

ways.

Thus invalid cases are

$$7 \cdot 7 = 49$$

Therefore,

$$462 - 49 = 413$$

So the required number is $413$.