How many integers n between 100 and 299 are there such that the hundre

Question

How many integers n between 100 and 299 are there such that the hundreds digit of n is less than the tens digit of n and the tens digit of n is less than the units digit of n ?

A. 48
B. 49
C. 50
D. 51
E. 52

ganand · Accepted Answer

Hi, We can divide the whole range into the two parts: 100 to 199 and 200 to 299. 100 to 199 Hundreds digit is fixed in this case,i.e. 1. Tens digit can be 2,3,4,5,6,7, and 8. For example, when tens digit is 2 we have the following possibilities: Number_counting1.png Number of ways = 1*1*7 = 7 ways. Similarly, when tens digit is 3 we have the following possibilities: Number_counting2.png Number of ways = 1*1*6 = 6 ways. Hence, the total number of ways = 7 + 6 + \dots + 1 = \frac{7 imes8}{2} = 28 numbers. 200 to 199 Hundreds number is 2. Tens digit can be 3,4,5,6,7, and 8. Hundreds digit can be 4,5,6,7,8, and 9. Hence, the total number in this case = 6 + 5 + \dots + 1 = \frac{6 imes7}{2} = 21 numbers. Total number = 28 + 21 = 49 numbers. Thanks.

Darshi04 · Answer

Solving using counting principle:

abc is a number in between 100 and 299, a is hundreds, b is tens and c is units place.
0<a<b<c<9
So, c is atleast 3.

✓c can take values between 3-9, inclusive = 7 values.
AND
✓b can take values between 2 and 8, inclusive (b is atleast 2) = 7 values.
AND
✓a can take 1 value = 1

Total possibilities = 7x7x1 = 49

Posted from my mobile device

Chethan92 · Answer

First number  N  such that the hundreds digit of n is less than the tens digit of n and the tens digit of n is less than the units digit of n is 123.
Then 124 ,125 ,126, 127, 128, 129 - Total 7 numbers.
Next is 134-139 - Total 6 numbers.
145-149 - 5 numbers.
156-159 - 4 numbers.
167-169 - 3 numbers.
178-179 - 2 numbers.
180 - 1 number.
total of 28 numbers between 100-200.

Similarly from 200 to 299.
234-239 - 6 numbers.
245-249 - 5 numbers.
256-259 - 4 numbers.
267-269 - 3 numbers.
278-279 - 2 numbers.
289 - 1 number.
total of 21 numbers.

Total is 28+21 = 49 numbers.

B is the answer.

Archit3110 · Answer

GMATinsight  :
Sir is there any other method to solve such question, where we dont end up counting every number?

N : (100A+10B+C)
so we need to find such specific no. ie 100A<10B<C

123,124,125,126,127,128,129,134,135,136.... so on..

this is a very time consuming method is there any short cut??

gracie · Answer

this is a triangular number sequence:
0,1,3,6,10,15,21,28...
here is the relationship between the hundreds block and number of compliant integers in each:
800-0
700-1
600-3
500-6
400-10
300-15
200-21
100-28
because the value for term n=n(n-1)/2,
the 7th term (200 block) will have 7*6/2=21 compliant integers
and the 8th term (100 block) will have 8*7/2=28 compliant integers
21+28=49 total compliant integers between 100 and 299
B

duynguyen612 · Answer

If the hundreds digit is 1, the tens digit has 7 cases, the units digit has 7 cases => has total 28 numbers
If the hundreds digit is 2, the tens digit has 6 cases, the units digit has 6 cases => has total 21 numbers 
--> has 49 numbers 
--> B

ritu1009 · Answer

Can you please explain how you get 28 and 21 numbers?

gulsam · Answer

It is faster and simpler, thanks! But there are possible signs <= And why a can take only 1 value? It is possible for a to be 1 and 2. Maybe you can explain this one?

Hamzee · Answer

Hi,

Can any one explain why a=1?

As we can see there are 2 digits 1 and 2 can be used ( 1 00- 2 99)?

Mintigate · Answer

It might some help

---------hundreds---------tens---------units--------- ------------1------------------2-----------2~9---------7choice ------------1------------------3-----------4~9---------6choice ------------1------------------4-----------5~9---------5choice ------------1------------------5-----------6~9---------4choice ------------1------------------6-----------7~9---------3choice ------------1------------------7-----------8~9---------2choice ------------1------------------8------------9----------1choice ---------hundreds---------tens---------units--------- ------------2------------------3-----------4~9---------6choice ------------2------------------4-----------5~9---------5choice ------------2------------------5-----------6~9---------4choice ------------2------------------6-----------7~9---------3choice ------------2------------------7-----------8~9---------2choice ------------2------------------8-------------9---------1choice Total 28 choices + 21 choices = 49 choices

AkmKawser · Answer

Just a typical Combination Problem. We can split the cases into two parts but remember it's not an option. It's a MUST. Cuz we have to find 100 to 299.100-199  AND  200-299. Thus we need to multiply these two cases rather than adding them up.

(1*7*7)/3! * (1*6*6)/3! = 7*7 = 49

How did I come up with this?

Case-1:100-199 
Hundreds Digit: 1 way (1)
Tens Digit : 7 ways ( 2,3,4,5,6,7,8)
Unit Digit : 7 ways (3,4,5,6,7,8,9)

Case-2:200-299 
Hundreds Digit: 1 way (2)
Tens Digit : 6 ways ( 3,4,5,6,7,8)
Unit Digit : 6 ways (4,5,6,7,8,9)

bumpbot · Answer

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How many integers n between 100 and 299 are there such that the hundre

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How many integers n between 100 and 299 are there such that the hundre

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