a ten - digit number is formed using the digits formed using the digits from zero to nine,every digits being used exactly once.,the probability that the number is divisible by 4 is

the total number of ways to form 10 digit number using digits 0 to 9 exactly once=10!-9!=10*9!-9!=(10-1)9!=9*9!

the number is divisible by 4, therefore the last two digits can be

 

04122032405260728092
08162436445664768496
  28 48 68 88 

but the digits are used only once , therefore we can not take 44 and 88 as the last digits.

now  when 0 is used in the last two digits,(i.e. 04,08,20,40,60,80=6 cases) 

the number of ways to fill remaining 8 places from 8 digits=8!

therefore the number of ways to form 10 digits number in the above 6 cases=6*8!

we have the remaining (24-2-6)=16 cases

so the number of ways (when 0 is not used in last two digits) to fill remaining 8 places with 8 digits(0 in cluded)

=8!-7!=(8-1)7!=7*7!

therefore the number of ways to form 10 digits number in the above 16 cases=16*(7*7!)=8*2*(7*7!)=14*8!

therefore the total number of required ways=6*8!+14*8!=(6+14)8!=20*8!

hence the required probability=

  • 7

no

  • -4

pls help me 2 solve this

  • 0
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