100! / 97!
= 100 * 99 * 98
= 970200
32C3 = 4960
10,000.
45
To find the number of three-digit combinations, we consider the digits from 000 to 999. Each digit can range from 0 to 9, giving us 10 options for each of the three digits. Therefore, the total number of three-digit combinations is (10 \times 10 \times 10 = 1,000).
5040, assuming none of the digits are the same. (Assuming they're not, there's 5040 unique combinations you can make out of 7 digits).
6
32C3 = 4960
Six combinations: 123, 132, 213, 231, 312, 321
Only 1. In a combination, the order of the digits does not matter.
10,000.
i guess the answer is 54. A group of three digits can be selected from the 5 digits in 9 ways and each group can be arranged in 3! (i.e., 6) ways respectively and hence the answer is 6*9=54
45
To find the number of three-digit combinations, we consider the digits from 000 to 999. Each digit can range from 0 to 9, giving us 10 options for each of the three digits. Therefore, the total number of three-digit combinations is (10 \times 10 \times 10 = 1,000).
5040, assuming none of the digits are the same. (Assuming they're not, there's 5040 unique combinations you can make out of 7 digits).
100C3= 100x99x98/3!= 161700 The notation is 100 choose 3
Assuming you are treating each number as a number and not as an individual unit, the numbers you can make from these digits are 899, 989 and 998.
017, 071, 107, 170, 701, 710. 6 combinations