IIT JEE , AIEEE Forum - Mathematics , Analytical Geometry

deep01

how many times does 8 occur when we list all the numbers from 1 to 1000??

(A)297 (B)300

(C)271 (d)273

pls also explain answer!!

master_purav

Its 300 times

Orloff

how? can you explain. plzz

shipra22_bhardwaj

master_purav

Lets split your problem as:

1-digit nos.:
Only once (in 8) ...1

2-digit nos.:
(i) 8 in units place ...9
(ii) 8 in tens place ...10

3-digit nos.:
(i) 8 in units place ...90
(ii) 8 in tens place ...90
(iii) 8 in hundreds place ...100

Thus 1+10+90+90+100=300...

master_purav

This ques is basically of Permutations and Combinations...

deep01

but if the limit extend upto 100000 the how can u do the same with ur trick

Orloff

Right !! But is any other method to this??????? Just asking... I mean by all the typical permutation and combination stuff?

master_purav

I cant think of any....

avinash.sharma

Numbers should be between 1 to 1000. It mean we shoud have to pay attention on arrangement of three digits as 1 itself is 001 and maximum number is 999.

lets we start our journey by thinking about single 8 so the cases may be

1. 8?? 2. ?8? 3. ??8 it mean for other two places we have (0,1,2,3,4,5,6,7,9) nine digits to arrange so total arrangments are ( 1 * 9 * 9) repeatitions are allowed. so total arrangements = 3 * ( 1*9*9) = 243 numbers

same way for repating 8 two times we have options 88? ?88 8?8 which yield the arrangemnts = 3 * (1*1*9) = 27 numbers

for three 888 we have only one number. Total numbers are 243+27+1=271

so right answer should be 271

rishikesh_anshu

the no can have either one 8 ,two 8 or three 8
case 1: three 8    888      1 way
case 2: two 8      ³C₂*9              ³C₂ for selecting 2 place for 8
                                                  and 9 values to fill third place
                                                 (excluding 8)
case 3:one eight    ³C₁*9*9    ³C₁ for selecting 1 place for 8
                                                  and 9 values to fill other place
                                                   (excluding 8)
Total= 1+³C₂*9 + ³C₁*9*9    =271

This is the general solution and can be extended further

dvrravi

yes , it's 271.
it's a typical problem btw..

priyesh

The question says how many times will eight occur & not how many nos. with 8 are possible therefore in the second case since 27 nos. are possible(having Two eights)

therefore no.of eights = 27 * 2=54

similarly in the third case only one no. 888 is possible having three eights

therefore total ways are 243 + (27*2) + (1*3)

=300

therefore no. of eights occuring are 300 whereas nos. possible are 271

master_purav

I agree with priyesh...

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