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How many ways are there to place the digits 0-9 in a sequence satisfying specific conditions?

a) 42
b) 90
c) 120
d) 362,880

User Tmullaney
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1 Answer

2 votes

Final answer:

The number of ways to place the digits 0-9 in a sequence satisfying specific conditions is given by 10 factorial (10!). The correct answer is d) 362,880.

Step-by-step explanation:

The number of ways to place the digits 0-9 in a sequence satisfying specific conditions can be determined by using the concept of permutations. In this case, there are 10 digits to be arranged in a sequence, which does not involve repetition. Therefore, the number of ways to arrange these digits is given by 10!, or 10 factorial.

The factorial of a number is the product of all positive integers less than or equal to that number. So, 10! = 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1. Evaluating this expression gives us a) 42, b) 90, c) 120, d) 362,880. Therefore, the correct answer is d) 362,880.

User Rohit Karadkar
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