To determine how many 4-digit numbers exist where all the digits are odd, we need to identify the possible choices for each digit position in a 4-digit number.
An odd digit can be one of the following numbers: 1, 3, 5, 7, or 9. Therefore, for any digit position, there are 5 possible choices.
Let's analyze each of the digit positions in the 4-digit number:
- Thousands Place: Since we're looking at a 4-digit number, the thousands place cannot be zero. It must be odd, and we have 5 choices (1, 3, 5, 7, 9).
- Hundreds Place: The hundreds place can also have any of those 5 odd numbers.
- Tens Place: Similarly, the tens place has 5 possibilities.
- Units Place: The units place also has 5 choices.
Using the basic principle of counting, the total number of 4-digit numbers with all digits odd is the product of the number of choices for each digit place.
5 \times 5 \times 5 \times 5 = 5^4 = 625
Thus, there are 625 4-digit numbers where all digits are odd. Therefore, the correct answer is: 625.
Similar Questions
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Important Questions from Permutation and Combination
On a chess board, in how many different ways can 6 consecutive squares be chosen on the diagonals along a straight path ?
There are 6 persons arranged in a row. Another person has to shake hands with 3 of them so that he should not shake hands with two consecutive persons. In how many distinct possible combinations can the handshakes take place ?
In a tournament of Chess having 150 entrants, a player is eliminated whenever he loses a match. It is given that no match results in a tie/draw. How many matches are played in the entire tournament?
The letters A, B, C, D and E are arranged in such a way that there are exactly two letters between A and E. How many such arrangements are possible?
There is a numeric lock which has a 3-digit PIN. The PIN contains digits 1 to 7. There is no repetition of digits. The digits in the PIN from left to right are in decreasing order. Any two digits in the PIN differ by at least 2. How many maximum attempts does one need to find out the PIN with certainty?
