find how many six-digit numbers can be formed from the digits 2,3,4,5,6,7 (with repetitions) if a) numbers formed must be even b) the numbers formed must be divisble by 25 c) the odd digits must occupy even position (2nd, 4th, 6th) and the even digits must occupy odd positions (1st, 3rd, 5th)
a)23328
b) 2592
c) 729
a) To find the number of six-digit numbers that can be formed using the digits 2, 3, 4, 5, 6, and 7, where the numbers must be even:
Step 1: Determine the restrictions on the first digit:
- The first digit cannot be zero, so you have 5 options (3, 4, 5, 6, 7).
Step 2: Determine the restrictions on the remaining five digits:
- For the remaining five digits, you have six options (2, 3, 4, 5, 6, 7) since repetitions are allowed.
Step 3: Multiply the number of options for each step:
- The number of possible six-digit numbers is found by multiplying the number of options at each step: 5 options for the first digit and 6 options for each of the remaining five digits.
- The total number of six-digit numbers that can be formed is: 5 x 6 x 6 x 6 x 6 x 6 = 64,800.
Therefore, there are 64,800 six-digit numbers that can be formed using the digits 2, 3, 4, 5, 6, and 7, where the numbers must be even.
b) To find the number of six-digit numbers that can be formed using the digits 2, 3, 4, 5, 6, and 7, where the numbers must be divisible by 25:
Step 1: Determine the restrictions on the last two digits:
- For the last two digits to form a number divisible by 25, the last digit must be 5.
- The second-to-last digit can be any of the digits 2, 3, 4, 6, or 7.
Step 2: Determine the restrictions on the remaining four digits:
- For the remaining four digits, you have five options (2, 3, 4, 6, 7) since repetitions are allowed.
Step 3: Multiply the number of options for each step:
- The number of possible six-digit numbers is found by multiplying the number of options at each step: 5 options for the second-to-last digit, 1 option for the last digit, and 5 options for each of the remaining four digits.
- The total number of six-digit numbers that can be formed is: 5 x 1 x 5 x 5 x 5 x 5 = 3,125.
Therefore, there are 3,125 six-digit numbers that can be formed using the digits 2, 3, 4, 5, 6, and 7, where the numbers must be divisible by 25.
c) To find the number of six-digit numbers that can be formed using the digits 2, 3, 4, 5, 6, and 7, where the odd digits must occupy even positions (2nd, 4th, 6th) and the even digits must occupy odd positions (1st, 3rd, 5th):
Step 1: Determine the restrictions on the odd positions (1st, 3rd, 5th):
- For the odd positions, you have three options (2, 4, or 6) since only even digits are allowed.
Step 2: Determine the restrictions on the even positions (2nd, 4th, 6th):
- For the even positions, you have three options (3, 5, or 7) since only odd digits are allowed.
Step 3: Multiply the number of options for each step:
- The number of possible six-digit numbers is found by multiplying the number of options at each step: 3 options for the first position, 3 options for the second position, and 3 options for the third position, giving you a total of 3 x 3 x 3 = 27 options.
Since the remaining three positions can be any digit (2, 4, 6, 3, 5, 7), we multiply the total number of options by 6 x 6 x 6 = 216.
- The total number of six-digit numbers that can be formed is: 27 x 216 = 5,832.
Therefore, there are 5,832 six-digit numbers that can be formed using the digits 2, 3, 4, 5, 6, and 7, where the odd digits must occupy even positions (2nd, 4th, 6th) and the even digits must occupy odd positions (1st, 3rd, 5th).
To find the number of six-digit numbers that can be formed from the digits 2, 3, 4, 5, 6, and 7, we can use the principles of combinatorics.
a) Numbers formed must be even:
An even number must end with either 2, 4, 6, or 7. So, in the last position, we have 4 choices.
For the remaining 5 positions, we have 6 choices for each position since we can use any of the 6 given digits.
Therefore, the total number of six-digit numbers that can be formed is 4 × 6^5 = 2,304.
b) Numbers formed must be divisible by 25:
For a number to be divisible by 25, it must end with the digits 25 or 75.
In this case, we have 2 choices for the last 2 positions.
For the remaining 4 positions, we still have 6 choices for each position.
Therefore, the total number of six-digit numbers that can be formed is 2 × 6^4 = 432.
c) Odd digits must occupy even positions and even digits must occupy odd positions:
For the odd digits (3, 5, 7), we can choose any one of them for the even positions (2nd, 4th, and 6th), which gives us 3 choices.
Similarly, for the even digits (2, 4, 6), we can choose any one of them for the odd positions (1st, 3rd, and 5th), which gives us 3 choices.
For the remaining 3 positions, we still have 3 choices for each position since we can use any of the 3 remaining digits.
Therefore, the total number of six-digit numbers that can be formed is 3^3 × 3^3 = 729.
Let me know if there is anything else I can help you with!
a) to be even, the unit digit must be even
so you have 3 evens
and since you can repeat any number, you will have
6*6*6*6*6*3 =
b) if a number is divisible by 25 it must end in 25, so what would be left over for the other spots?
c) so the number must look like this:
EOEOEO , fill in the spaces.