Question

How many three digit positive integers have their hundreds tens and units digits in ascending order?

Answer 1

Answer:

there are Six three-digit positive integers that can be formed from the digits 3 , 4 , and 8 . In fact, we could list them all if we really wanted to: 348 , 384 , 438 , 483 , 834 , and 843 . The correct answer is B, 6 .Therefore, there are a total of 1 + 3 + 6 + 10 + 15 + 21 + 28 + 36 = 120 three-digit descending numbers. Since there are 999 - 100 + 1 = 900 three-digit numbers, the probability of picking a descending number is 120/900 = 12/90 = 2/15

Answer 2

84 integers

Step-by-step explanation:

Three digit positive integers have their hundreds tens and units digits in ascending order

as the three digit numbers are start from 100

then from (100- 200)

the numbers are 123-129 total (7)

134-139 total(6)

145-149 total(5)

156-159 total(4)

167-169 total (3)

178-179 total (2)

189(1)

then on adding 7+6+5+4+3+2+1 = 28 numbers

if we continue the same upto 999

we get

28+21+15+10+6+3+1 = 84 numbers

hence ,

there are total 84 integers

#Learn more:

How many three-digits numbers can be generated from 1,2,3,4,5,6,7,8,9 such that the digits are in ascending order ?

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