Let A be event that there is no zero and B be event that there is no one.

We need n(A'' intersect B'') or n (A union B)''

Use De Moivre''s theorem.

hi shaleen,

atleat one 1 and one 0 makes 3 cases

1) one 0 and one 1

then first we will select 2 out of 4 places for 0,1 and place 8 digits in remaining 2

we get C(4,2)*8^2 and the can be arranged in 2 ways so whole expression becomes

2*C(4,2)*8^2 = 768

2)2 0''S and 2 1''s

so we have to arrange them amongst each other 4!/(2!.2!) = 6

3)2 0''s and one 1

first we select 3 places out of 4 then place 8 digits in one place and arrange 2 0''s and 1 one amongst themseleves

we get the expression as C(4,3)*(3!/2!)*8

similar will be the case when there will be 2 1''s and one 0

so 2*C(4,3)*(3!/2!)*8=192

4) 3 0''s and one 1

we just have to arrange them we can do it by 4!/(3!*1!)

and similar will be the case when there are 3 1''s and one 0

so we get final expression as 4*2=8

so when we add 768+192+6+8 = 974

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