can someone correct this for me parts A-E and if i am wrong can you help me....please
How many 7-digit telephone numbers are possible if the first digit cannot be zero and.....
a) only odd digits may be used?
my answer: 8125
b) the telephone number must be multiple of 10(that is , it must end in zero)?
my answer: 1,000,000
c) the telephone number must be a multiple of 100?
my answer 100,000
d) the first 3 digits are 481?
my answer 10,000
e) no repetitions are allowd?
my answer: 120
I will be happy to critique your work.
Sure, let's go through each part one by one:
a) Correct! If only odd digits may be used, there are only 5 options for each digit (1, 3, 5, 7, 9). Since we have 7 digits, the total number of possibilities is 5^7 = 78125.
b) Incorrect. If the telephone number must be a multiple of 10 (ending in zero), we need the last digit to be zero. The rest of the digits can be any digit from 0 to 9, including zero for the first digit. Therefore, the total number of possibilities is 10^6 = 1,000,000.
c) Incorrect. If the telephone number must be a multiple of 100, we need the last two digits to be zero. The first digit cannot be zero, so the rest of the digits can be any digit from 0 to 9. Therefore, the total number of possibilities is 9 * 10^5 = 900,000.
d) Incorrect. If the first 3 digits are 481, we have fixed values for those digits. The remaining 4 digits can be any digit from 0 to 9. Therefore, the total number of possibilities is 10^4 = 10,000.
e) Incorrect. If no repetitions are allowed, the first digit can be any digit from 1 to 9. The second digit can be any digit from 0 to 9, excluding the digit used for the first digit. Similarly, the third digit can be any digit from 0 to 9, excluding the digits used for the first and second digits. Continuing this pattern, the total number of possibilities is 9 * 9 * 8 * 7 * 6 * 5 * 4 = 326,592.
I hope this helps! Let me know if you have any further questions.