# 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

 04 12 20 32 40 52 60 72 80 92 08 16 24 36 44 56 64 76 84 96 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= • 12

no

• -5

pls help me 2 solve this

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