1.how many different 4-digit even number can be formed from the gigits 1,3,5,6,8 and9 if no repetition of digits allowed?
the number must end in 8 or 6.
In either case, there are 5*4*3 = 60 possible numbers
so, 120 altogether.
To determine the number of different 4-digit even numbers that can be formed from the digits 1, 3, 5, 6, 8, and 9 with no repetition of digits allowed, we can follow these steps:
Step 1: Count the number of choices for the thousands place.
Since the numbers must be even, the only possible choice for the thousands place is 8. Therefore, there is only 1 choice for the thousands place.
Step 2: Count the number of choices for the hundreds place.
Since we have used one number already in the thousands place, there are 5 remaining numbers: 1, 3, 5, 6, and 9. To ensure the number is even, we need to exclude the number 1. Therefore, there are 4 choices for the hundreds place.
Step 3: Count the number of choices for the tens place.
At this point, we have used two numbers already (8 and 1), leaving us with 4 remaining numbers: 3, 5, 6, and 9. We need to exclude the number 1 and choose an even number. Thus, there are 2 choices for the tens place.
Step 4: Count the number of choices for the ones place.
Since we have used three numbers already (8, 1, and the even number chosen for the tens place), there are 3 remaining numbers: 3, 5, and 9. We can choose any of these numbers for the ones place. Therefore, there are 3 choices for the ones place.
Step 5: Multiply the number of choices at each step.
To find the total number of different 4-digit even numbers, we need to multiply the total number of choices at each step. Thus, 1 choice for the thousands place multiplied by 4 choices for the hundreds place, multiplied by 2 choices for the tens place, multiplied by 3 choices for the ones place.
1 x 4 x 2 x 3 = 24
Therefore, there are 24 different 4-digit even numbers that can be formed from the given digits with no repetition allowed.
To find the number of different 4-digit even numbers that can be formed from the given digits (1, 3, 5, 6, 8, and 9) without repetition, we need to consider a few conditions:
1. The thousands digit cannot be zero, as 4-digit numbers cannot start with zero.
2. The units (ones) digit must be an even number, i.e., either 6 or 8.
Now let's break down the problem into steps:
Step 1: Determine the number of choices for each digit.
- Thousands digit: Since zero is not allowed, we have 5 options: 1, 3, 5, 6, 8.
- Hundreds digit: We have 5 digits remaining after choosing the thousands digit.
- Tens digit: We have 4 digits remaining after choosing the thousands and hundreds digits.
- Units (ones) digit: We have only 2 options, either 6 or 8, as it must be even.
Step 2: Multiply the number of choices for each digit.
To get the total number of possible combinations, we multiply the number of choices for each digit.
Number of choices for the thousands digit = 5
Number of choices for the hundreds digit = 5
Number of choices for the tens digit = 4
Number of choices for the units (ones) digit = 2
Total number of possible combinations = 5 * 5 * 4 * 2 = 200
Therefore, there are 200 different 4-digit even numbers that can be formed from the given digits without repetition.