The number of four-digit natural numbers whic | vedclass

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Question

(A) To form a four-digit number with exactly two distinct digits,we consider two cases:Case $1$: The digit $0$ is not included.We choose $2$ non-zero

Vedclass · Accepted Answer

$567$

Vedclass · Answer

(A) To form a four-digit number with exactly two distinct digits,we consider two cases:Case $1$: The digit $0$ is not included.We choose $2$ non-zero digits from ${1, 2, ..., 9}$ in $^9C_2$ ways. Each of these $2$ digits can be placed in $4$ positions such that each digit appears at least once. The number of ways to arrange them is $2^4 - 2 = 14$. So,$^9C_2 	imes 14 = 36 	imes 14 = 504$.Case $2$: The digit $0$ is included.We choose $1$ non-zero digit from ${1, 2, ..., 9}$ in $^9C_1$ ways. The number must start with the non-zero digit. For the remaining $3$ positions,we must choose at least one $0$ and at least one of the chosen non-zero digit. The number of ways to fill the remaining $3$ positions is $2^3 - 1 = 7$. So,$^9C_1 	imes 7 = 9 	imes 7 = 63$.Total numbers = $504 + 63 = 567$.

The number of four-digit natural numbers which contain exactly two distinct digits is:

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