Find the number of eight digit numbers in which the fifth digit is numerically greatest and the numbers to the left and right of it are in decreasing order (not strictly decreasing)... (repetition allowed)...
106if 5th digit is 2, then no. of ways are 1*2*2*1*2*2*2*2=64
If 5th digit is 3, then no. of ways are 2*3*3*1*3*3*3*3
If 5th digit is 4, then no. of ways are 3*4*4*1*4*4*4*4
If 5th digit is 5, then no. of ways are 4*5*5*1*5*5*5*5
If 5th digit is 6, then no. of ways are 5.6.6.1.6.6.6.6
If 5th digit is 7, then no. of ways are 6.7.7.1.7.7.7.7
If 5th digit is 8, then no. of ways are 7.8.8.1.8.8.8.8
If 5th digit is 9, then no. of ways are 8.9.9.1.9.9.9.9
total is sum of these
1@ Asish
I nthe first case fifth digit is 2 you have included 10020011, 10020101 and 10120001 etc. in accordance to the number of ways. This violates the condition in the question.
There such numbers in every set!!!
Further, digits are counted from the right, so it is only 3 numbers which cannot be 0
Which i dont know as to which way U took
So in yer method answer should have been 1*1*1*1*2*2*2*
So you have considered many numbers that do not fit into the conditions.
I could have answered this if I knew how many[2]
1@asish... explain ur method plz... i dun have answer to dis one...
infact, i didn't copy dis question directly, came across a similar question, added some extra flavour to it [by adding (not strictly increasing) ] .......
so post ur complete solution plz...
106@kalyan... i took the position of 2 backwards.
@ MAK
case (i) 5th digit is two
So, 1st digit is 1 2nd digit is 0,1 ... till 4th and 6th to 8th... so. it is 1*2*2*2*1*2*2*2
case (ii) 5th digit is 3
So, 1st digit is 1,2 2nd to 4th and 6th to 8th can be 0,1,2 ..
So it is 2*3*3*3*1*3*3*3
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for all cases... u can see what i have written here... is in the reverse order of what i had written there .. as i was counting backwards from 8th digit to 1st
1well, nyc method... a slight modification is required... even d no. 11111111 satisfies d given condition... so u must add one to all ur cases
then it would be... 1*2*2*2*1*2*2*2 + ...... 9*10*10*10*1*10*10*10...
remember i said not strictly decreasing...
anywayzzz... nyc solution... :)
1[46]
u r rite... i'm sorry...!!!
actually i had that thought in my mind while posting d question but couldn't present it properly, in hurry...
well, ur answer is correct... :)
1asish tere 1st case mein 5th digit is 2 , isme ek number aisa bhi banega jisme 2nd digit 0 and 3rd digit 1 hogi which voilates the rule