100C3= 100x99x98/3!= 161700 The notation is 100 choose 3
-2
-8
32C3 = 4960
100! / 97! = 100 * 99 * 98 = 970200
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).
6
-2
-8
32C3 = 4960
Six combinations: 123, 132, 213, 231, 312, 321
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
100! / 97! = 100 * 99 * 98 = 970200
Only 1. In a combination, the order of the digits does not matter.
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).
There are a total of 1,000 three-digit combinations from 000 to 999. This includes all combinations where the digits can range from 0 to 9, allowing for repetitions. Each of the three digit positions can have 10 possible values (0-9), leading to (10 \times 10 \times 10 = 1,000) combinations.
45
10,000.