Question

number of ways we can form a number of 4 digits that have atmost 2 different digits

Answer 1

576

Answer 2

To find the number of 4-digit numbers that have at most 2 different digits, we consider two cases:

Case 1: The number has exactly 1 different digit.

This means all four digits are the same. Let the number be $AAAA$. Since it is a 4-digit number, the first digit $A$ cannot be 0. So, $A$ can be any digit from 1 to 9. Possible numbers are 1111, 2222, ..., 9999. Number of ways = 9.

Case 2: The number has exactly 2 different digits.

Let the two distinct digits be $d_1$ and $d_2$. The 4-digit number must contain both $d_1$ and $d_2$. We need to consider two subcases based on whether 0 is one of the digits.

Subcase 2.1: One of the distinct digits is 0.

Let the two distinct digits be $x$ and $0$, where $x \in \{1, 2, ..., 9\}$. There are 9 choices for $x$. The 4-digit number must contain both $x$ and $0$. Since the number is a 4-digit number, its first digit cannot be 0. Therefore, the first digit must be $x$. So the number is of the form $xBCD$. The remaining three digits ($B, C, D$) can be either $x$ or $0$. There are $2^3 = 8$ possible arrangements for $BCD$. However, we must ensure that $0$ is present among $B, C, D$. The only arrangement that does not contain $0$ is $xxx$. So, we exclude the case $xBCD = xxxx$ (which would fall under Case 1). Number of valid arrangements for $BCD$ is $2^3 - 1 = 7$. For each of the 9 choices for $x$, there are 7 ways to form the number. Number of ways = $9 \times 7 = 63$.

Subcase 2.2: Neither of the distinct digits is 0.

Let the two distinct digits be $x$ and $y$, where $x, y \in \{1, 2, ..., 9\}$ and $x \neq y$. The number of ways to choose two distinct digits from $\{1, 2, ..., 9\}$ is $\binom{9}{2}$. $\binom{9}{2} = \frac{9 \times 8}{2} = 36$. For each chosen pair $\{x, y\}$, we need to form a 4-digit number using only $x$ and $y$, such that both $x$ and $y$ are present. Since neither $x$ nor $y$ is 0, the first digit can be either $x$ or $y$. Each of the four positions in the number can be filled with either $x$ or $y$. So there are $2^4 = 16$ total arrangements. From these 16 arrangements, we must exclude the cases where only $x$ is used ($xxxx$) and where only $y$ is used ($yyyy$). These cases would fall under Case 1. So, the number of valid arrangements for each pair $\{x, y\}$ is $2^4 - 2 = 14$. Number of ways = $\binom{9}{2} \times 14 = 36 \times 14 = 504$.

Total number of ways:

Sum of ways from Case 1, Subcase 2.1, and Subcase 2.2.

Total = $9 + 63 + 504 = 576$.