jitender lakhanpal
Last Activity: 11 Years ago
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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Jitender
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