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Stat v. 62

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Can any one here tell me why u cnt do
2C2*4C2*9C2*9C6
In quest 6 part i ???? :( :'(

No, there were to be 3 such options made. 2 was the least it had to be and the total was 12. There were three groups so it (2C2*4C2*9C8)+(2C2*4C3*9C7) +(2C2*4C4*9C6)
 
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No, there were to be 3 such options made. 2 was the least it had to be and the total was 12. There were three groups so it (2C2*4C2*9C8)+(2C2*4C3*9C7) +(2C2*4C4*9C6)
I am nt sayng to justify ur ans. I am askng why my method is wrng?
 
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I am nt sayng to justify ur ans. I am askng why my method is wrng?

Right there in the answer is the reason. Its like you need to chose 12 out of these and there must be atleast 2 of all three kinds, to begin with take 2 of the 2 of oen kind, 2 of the four of the other and to make it twelve 8 of the 9 of the third batch i.e. 2C2*4C2*9C8. Moving on you can make other options but not getting any lower than 2 of each type so 2C2*4C3*9C7 Now you cant have more than 4 of the second batch so third and last option 2C2*4C4*9C6 There have to be atleast 2 of each so you consider all options above atleast 2 in each batch. Hope you get it now :)
 
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Right there in the answer is the reason. Its like you need to chose 12 out of these and there must be atleast 2 of all three kinds, to begin with take 2 of the 2 of oen kind, 2 of the four of the other and to make it twelve 8 of the 9 of the third batch i.e. 2C2*4C2*9C8. Moving on you can make other options but not getting any lower than 2 of each type so 2C2*4C3*9C7 Now you cant have more than 4 of the second batch so third and last option 2C2*4C4*9C6 There have to be atleast 2 of each so you consider all options above atleast 2 in each batch. Hope you get it now :)
Sory still no !
 
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