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Bumek [7]
3 years ago
5

In a three-digit positive integer , if the hundreds digit cannot be 1 and the neighbor digits cannot be repetition, how many pos

sibilities of these integers? A. 729 B. 504 C. 576 D. 448 E. 648
Mathematics
1 answer:
andreev551 [17]3 years ago
6 0

Answer:

Option (E) 648

Step-by-step explanation:

the 3 digit number can be represented by the blanks as " _ _ _  "

Now,

we have 10 choices ( i.e 0,1,2,3,4,5,6,7,8,9) available for each place in the blank if no condition is applied.

For the hundreds digit, using the conditions given in the question, we have 8 choices left

as 1 and 0 cannot be included in the hundreds place.

for the tens place

we will have 9 choices left out of 10 ( as 1 choice is less because we cannot have same number as on the hundred place )

similarly, for the ones place we have 9 choices left out of 10 ( as 1 choice is less because we cannot have same number as on the tens place )

Therefore,

Total possibilities = 8 × 9 × 9 = 648

Hence,

Option (E) 648

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