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In an examination of 9 papers a candidate has to pass in more papers than the number of papers in which he fails in order to be successful.the number of ways in which he can be unsuccesful is ?

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let if a student fails in n subjects where n <=9.then he must pass in atlest n+1 subjects.

so, tolal no. of successful ways= c(9,n+1) +c(9,n+2) + .......+c(9,8) + c(9,9). =s

so, toatal no. of unsuccesful ways=2^9 - s

=2^9-(2^9-∑c(9,n-k)) [where k is from 0 to n]

=∑c(9,n-k) ans.

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