IMO D
one digit number-1
two digit- 7X or X7
7X= 10 times
X7- 9times

thee digits-7XX,X7X,XX7
7XX- 100 times
X7X=90 times
XX7-90 times
total 1+10+9+100+90=90=300

thanks. Yup 300 is the answer .. could you explain how you are calculating the thee digits-7XX,X7X,XX7

7XX- 100 times
X7X=90 times
XX7-90 times

Also, were you able to do the question in under 2 mins ?

thanks

Ad_Astra_Per_Aspera wrote:thanks. Yup 300 is the answer .. could you explain how you are calculating the thee digits-7XX,X7X,XX7

7XX- 100 times
X7X=90 times
XX7-90 times

Also, were you able to do the question in under 2 mins ?

thanks

yes I was able to do it within a minute but that might be because I got the underlying idea as soon as I saw the question
now
7XX- x can take any values from 0-9. so total number of ways=10(for tenths digit)*10(for unit's digit)=100

X7X- here the X at the hundred's place can only take values from 1 to 9
so total number of ways=9*10=90
same for XX7

gotcha .... thanks.

im scared of these kind of questions because i always worry that my counts would overlap. like in this question, while 7X can appear 10 times (70-79), how can X7 still appear 9 times (17,27, 37...97) as "77" is overlapped!? am i thinking too much?

jkpimp168 wrote:im scared of these kind of questions because i always worry that my counts would overlap. like in this question, while 7X can appear 10 times (70-79), how can X7 still appear 9 times (17,27, 37...97) as "77" is overlapped!? am i thinking too much?

ok thats the trick..
we are not being asked how many numbers..we are being asked how many times. so You gotta count those numbers twice:))

if you want to be on the safe side, try this:
numbers from 0-999 can be represented as XXX, 0<=x<=9
only one 7- like 7XX where x is any value from 0-9 except 7
3*9*9=243
two 7's=3*9=27. now each number contains two 7's. so total 7's=54:)
three 7's. only one number 777. But, three 7's
total 7's=243+54=3=300.

tohellandback wrote:

jkpimp168 wrote:im scared of these kind of questions because i always worry that my counts would overlap. like in this question, while 7X can appear 10 times (70-79), how can X7 still appear 9 times (17,27, 37...97) as "77" is overlapped!? am i thinking too much?

ok thats the trick..
we are not being asked how many numbers..we are being asked how many times. so You gotta count those numbers twice:))

if you want to be on the safe side, try this:
numbers from 0-999 can be represented as XXX, 0<=x<=9
only one 7- like 7XX where x is any value from 0-9 except 7
3*9*9=243
two 7's=3*9=27. now each number contains two 7's. so total 7's=54:)
three 7's. only one number 777. But, three 7's
total 7's=243+54=3=300.

ooo...ook....i think i got it. so in "77", 7 appears 2 times, and 3 times in "777", yea? thx so much!

jkpimp168 wrote:

tohellandback wrote:

jkpimp168 wrote:im scared of these kind of questions because i always worry that my counts would overlap. like in this question, while 7X can appear 10 times (70-79), how can X7 still appear 9 times (17,27, 37...97) as "77" is overlapped!? am i thinking too much?

ok thats the trick..
we are not being asked how many numbers..we are being asked how many times. so You gotta count those numbers twice:))

if you want to be on the safe side, try this:
numbers from 0-999 can be represented as XXX, 0<=x<=9
only one 7- like 7XX where x is any value from 0-9 except 7
3*9*9=243
two 7's=3*9=27. now each number contains two 7's. so total 7's=54:)
three 7's. only one number 777. But, three 7's
total 7's=243+54=3=300.

ooo...ook....i think i got it. so in "77", 7 appears 2 times, and 3 times in "777", yea? thx so much!

yepp!!!

The position of no. 7 ( units,tens,hundreds) can be chosen in 3C1 ways
the other 2 places can filled by all the 10 digits ( 0 to 9)

Total no.s = 3C1*10*10 = 300