Latest Elitmus Aptitude Question SOLUTION: How many 3 digit numbers less than 1000 are there in which a 3 is followed by a 7 ?

ans. 19
first : _37
blank can be filled with (1,2,3,4,5,6,7,8,9)
hence 9 ways

second : 37_
blank can be filled with (1,2,3,4,5,6,7,8,9)
hence 10 ways

total = 9+10 = 19.....

Answer will be 463.. you have not posted d question correctly, its lyk how many 3 digits no. are dere in which if dere is a 3 it is followed by a 7.. that simply doesnt mean dat we shall not count d no. without 3.

since 3 iz fallowed by a 7 so 37 r always togeother.now if 37_ then last place can be filled in 10 ways.
and if _37 then first place can be filled in 9 ways so total no. 10+9=19
ans.19

Answer will be 463
suppose d 3 digit no. starts with 1
10__(This blank space can b filled in 7 ways 2,4,5,6,7,8,9)
in d above case we cannot take 3 since it should be followed by 7 which is not possible here.
now we take-
11__ this space can also be filled in 7 ways
12__ 7 ways
13__ this space can be filled only by 1 way and dat is 7.
14__ again 7 ways
15__ 7 ways
16__ 7 ways
17__ 7 ways
18__ 7 ways
19__ 7 ways
that gives us 57 ways in total

We will follow d same procedure for 3 digits starting with 2,4,5,6,8,9(but not in case of 3 and 7)
thus we get 57*7= 399 ways

now lets check condition for 3. 3 is always followed by 7 and nothing else
thus
37__ this space can be filled in 8 ways(0,1,2,4,5,6,8,9)
no other condition is possible in case of 3.
that makes our total to be 399+8=407

now lets check condition for 7
70__ this can be filled in 7 ways
71__ 7 ways
72__ 7 ways
73__ this condition is not possible since 3 should be followed by 7 but we cannot repeat 7.
74__ 7 ways
75__ 7 ways
76__ 7 ways
78__ 7 ways
79__ 7 ways
this gives us 56 ways
therefore our total becomes 407+56=463 and this is the answer.

the no. can be of the form 37_ or _37.
in case of 37_ the empty space can b filled up by no. 0,1,2,4,5,6,8,9 i.e. 8 ways
in case of _37 the space can be filled up by any of the above mentioned digit except 0(as its a 3 digit no.)thus no. of ways=7
thus total no. of ways=7+8=15

--- in this first two digit is 3 and 7 and last digit can be placed in 10 different ways (0123456789) as there is nothing given about repitation so we can take repitation.again --- first digit can be filled in 9 ways so, total 10+9=19 digits

37- 0,1,2,4,5,6,7,8,9---->9
-37 1,2,4,5,6,7,8,9------>8
9+8=17

37_ last digit can be filled by (1,2,4,5,6,8,9) total 7 ways.
_37 first digit can also be filled by (1,2,4,5,6,8,9) total 7 ways.
answer is 7+7=14

there was something more in the question that no digits being repeated so total there are 19 and if we dont take 737 and 337 it will be 17 and options were 14 17 463 and 520

I am also getting the answer 15 but there were 14 17 and two other options in paper around 463 and 520

thing will look like 37_ or _37. hence total no. of ways will be 2*10c1 -1=19 ways where 037 is not a possible 3 digit number

No. of 3 digits possible in which 3 is followd by 7 are:
37_
3_7
_37

Now 1st can be filled in 10 ways(0-9).
2nd can be filled with 10 ways.
3rd 1 can be filled with 9 way(as 0 will make it 2 digit nos.).
Hence, 10+10+9=29
answer shd be 29 digits possible.

as 3 digit nmber so nos which dont have a 3 ,may be made up of (0,1,2,4,5,6,7,8,9)toal digits =9 ,out of which at 100th palace we cant use 0.so it is 8*9*9=648(as question asked no. in which a 3 is followed by a 7,however it didn,t ask for a 7 preceded by a 3,so we considered 7)now 648+19=667