CSIR JUNE 2011 PART B QUESTION 22 SOLUTION (SUBSCRIBE MY BLOG TO GET REGULAR UPDATES)

The number of 4 digit numbers with no two-digit common is
1. 4536,
2. 3024,
3.5040,
4. 4823.
Solution: We will explicitly calculate the number of four-digit numbers with no two-digits common (repeated).

The first digit can be any number between $0$ to $9$ except it cannot be $0$. Because if the first digit is zero then it is no more a four-digit number. Hence the first digit has $9$ possibilities $1$ to $9$.

The second digit can be any number between $0$ to $9$  except the one used in the first digit. But this digit can be 0. Hence the possibilities are $\{0,1,2,\dots,9\} \backslash \{first digit\}$ hence there are $9$ possibilities.

The third digit can be any number between $0$ to $9$  except the numbers used in the first and the second digit. Hence the possibilities are $\{0,1,2,\dots,9\} \backslash \{first digit, second digit\}$ hence there are $8$ possibilities. Note that, by our choice first and the second digit was different.

The fourth digit can be any number between $0$ to $9$  except the numbers used in the first, second and third digit. Hence the possibilities are $\{0,1,2,\dots,9\} \backslash \{first digit, second digit, third digit\}$ hence there are $7$ possibilities.

Hence the total number of possibilities is equal to 9 x 9 x 8 x 7 = 4536!!!.


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