How many ways can the digits 0 through 9 be arranged if each digit can be used only once?
a) 1
b) 9! 1
c) 0!

The number of times that 0 to 9 can be arranged if each digit is used once is c. 10!

There are 10 numbers in the range of 0 - 9 which means that after the first number is selected to be arranged, there will be 9 numbers left to be arranged and then 8 and so on till all numbers are ordered.

The number of ways these numbers can be arranged is therefore:

= 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1

This is the same as 10!.

In conclusion, the numbers can be arranged in 10! ways.

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Hence,  the number of ways can the digits 0 through 9 be arranged if each digit can be used only once is:

10!=3628800

Step-by-step explanation:

The numbers of ways in which we can arrange the digits 0-9 in such a way that each digit can be used only once is:

10!

( since the total number of digits from 0-9 is 10

so we have to arrange these 10 digits in order such that repetition is not allowed

so let in first place we can have any of the 10 digits.

then in second place we can choose out of the remaining 9 digits.

for third place we can choose any of the remaining 8 digits and so on till the final place so the arrangement will be:

10×9×8×7×6×5×4×3×2×1=10!=362800

)

s + (a + 178) = 2000

step-by-step explanation:

2000 - 178 = 1822

1822 / 2 = 911

911 + 178 = 1089

1089 + 911 = 2000

abha = 1089

shane = 911

the probability is 0