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A “descending number” is a three-digit number, such that the
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17 Oct 2017, 22:22
17 Oct 2017, 22:22
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A “descending number” is a three-digit number, such that the units digit is less than the tens digits and the tens digit is less than the hundreds digit. What is the probability that a three-digit number chosen at random is a “descending number”?
(A) 3/25
(B) 2/9
(C) 2/15
(D) 1/9
(E) 1/10
Kudos for correct solution.
(A) 3/25
(B) 2/9
(C) 2/15
(D) 1/9
(E) 1/10
Kudos for correct solution.
Re: A “descending number” is a three-digit number, such that the
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18 Oct 2017, 05:50
18 Oct 2017, 05:50
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Three-digit numbers are 999-100+1 = 900.
For any arrangement of three numbers, there is only one in which the numbers are arranged so that the units are less than the tens, which are less than the hundreds: 10x9x8 = 720 are the total different combinations of three different integers. Only one of the six possible arrangements (3! = 6) matches up with the restraints in the problem. Thus, 720/6 = 120.
Then, 120/900, which simplifies to 2/15. Answer C
For any arrangement of three numbers, there is only one in which the numbers are arranged so that the units are less than the tens, which are less than the hundreds: 10x9x8 = 720 are the total different combinations of three different integers. Only one of the six possible arrangements (3! = 6) matches up with the restraints in the problem. Thus, 720/6 = 120.
Then, 120/900, which simplifies to 2/15. Answer C
Re: A “descending number” is a three-digit number, such that the
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18 Oct 2017, 23:42
18 Oct 2017, 23:42
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Bunuel wrote:
A “descending number” is a three-digit number, such that the units digit is less than the tens digits and the tens digit is less than the hundreds digit. What is the probability that a three-digit number chosen at random is a “descending number”?
(A) 3/25
(B) 2/9
(C) 2/15
(D) 1/9
(E) 1/10
Kudos for correct solution.
(A) 3/25
(B) 2/9
(C) 2/15
(D) 1/9
(E) 1/10
Kudos for correct solution.
The total 3 digit numbers can be formed by FCP i.e = 9*10*10 (we have digit from 0 -9 and there are no restriction then the hundred digit can be from 1-9,as it cannot start from 0 , tens digit can be 0-9 and units can also be 0-9).
Now we have to choose the three digit numbers in descending order -
We can leave out 100 - 200 = as they donot form in descending order
Next from 200 to 200 = we have only one number =210
Next from 300 to 400 = we have 3 nos = 310,320,321
Next from 400 to 500 = we have 6 nos = 410,420,421,430,431,432
Now we find that it is following a pattern and i.e
200 = 1
300 = 3 (2+1 from previous possible ways , 2 is the hundred unit )
400 = 6 (3+3)
500 = 10 (4+6,similarly 4 is the hundred digit and 6 in the value from the previous)
600 = 15(5+10)
700 = 21 (6+15)
800 = 28 (7+21)
900 = 36(8+28)
Therefore no. of possible ways = 36 + 28 + 21 + 15 + 10 + 6 + 3 + 1 =120
Therefore probability = \(\frac{120}{900} = \frac{2}{15}\)
