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how many 7-digit numbers are there where all the digits are different and in increasing order? for example, the numbers 1345689 and 2456789 would be counted. (the first digit cannot be zero.) (a) 20 (b) 24 (c) 28 (d) 32 (e) 36

Sagot :

bharathparasad577

There are 544,320 7-digit numbers that are all unique and arranged in increasing order.

We apply the Fundamental Counting Principle to this issue. You are aware that a number has seven digits. According to this theory, you must multiply the potential integers for each digit. There are 9 potential numbers if the initial number can't be zero. Consequently, the first digit's value is 9. Any number, but less than 1, could be used for the second digit since it was already used in the first. Thus, 10 – 1 would equal 9. The value in the third digit must be less than 1 and in the second digit. In this case, 9 – 1 Equals 8. And the list goes on.

Therefore, the number of seven-digit numbers in which all digits are different and in increasing order is:

9 × 9 × 8 × 7 × 6 × 5 × 4 = 544,320

Hence, there are 544,320 7-digit numbers with all different digits in increasing order.

Learn more about increasing order here:

https://brainly.com/question/3844889

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