How many different 5 digit numbers can be formed from the digits 9, 0, 4, 1, 6?
96
Calculating Different 5-Digit Numbers
The problem asks us to find the total number of distinct 5-digit numbers that can be formed using the digits 9, 0, 4, 1, and 6, with each digit used exactly once in each number.
A 5-digit number cannot have 0 in the first position (the ten thousand's place). The given digits are 9, 0, 4, 1, 6.
Understanding the Constraints
- We are forming a 5-digit number.
- The digits available are 9, 0, 4, 1, 6.
- Each digit must be used exactly once.
- The first digit (at the ten thousand's place) cannot be 0.
Step-by-Step Calculation Method
Let's consider the positions in the 5-digit number from left to right:
- First Position (Ten Thousands Place): This position cannot be 0. The available digits are 9, 4, 1, and 6. So, there are 4 choices for the first position.
- Second Position (Thousands Place): After placing a digit in the first position, we have 4 digits remaining (including 0, as it is now allowed). So, there are 4 choices for the second position.
- Third Position (Hundreds Place): After placing digits in the first two positions, we have 3 digits remaining. So, there are 3 choices for the third position.
- Fourth Position (Tens Place): After placing digits in the first three positions, we have 2 digits remaining. So, there are 2 choices for the fourth position.
- Fifth Position (Units Place): After placing digits in the first four positions, we have 1 digit remaining. So, there is only 1 choice for the fifth position.
To find the total number of different 5-digit numbers, we multiply the number of choices for each position:
Total numbers = (Choices for Pos 1) $\times$ (Choices for Pos 2) $\times$ (Choices for Pos 3) $\times$ (Choices for Pos 4) $\times$ (Choices for Pos 5)
Total numbers = $4 \times 4 \times 3 \times 2 \times 1$
Total numbers = $16 \times 6$
Total numbers = $96$
Alternative Permutation Method
We can also calculate this using the concept of permutations.
The total number of permutations of 5 distinct digits (if there were no restriction on the first digit) is $5!$ (5 factorial).
$5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$
Now, we need to subtract the number of arrangements where 0 is in the first position (which are not valid 5-digit numbers). If 0 is in the first position, we need to arrange the remaining 4 digits (9, 4, 1, 6) in the remaining 4 positions. The number of ways to do this is $4!$ (4 factorial).
$4! = 4 \times 3 \times 2 \times 1 = 24$
The number of valid 5-digit numbers is the total number of permutations minus the number of permutations with 0 in the first position.
Number of 5-digit numbers = Total permutations - Permutations with 0 in the first position
Number of 5-digit numbers = $120 - 24$
Number of 5-digit numbers = $96$
Both methods yield the same result.
Therefore, 96 different 5-digit numbers can be formed from the digits 9, 0, 4, 1, 6.
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?
