341What has this question got to do with permutation?
A hint: Start by answering how many students got exactly k wrong answers. Then how many misfired on exactly (k-1) questions and work your way down
in a certain test ,ai students gave wrong answers to at least i questions where i=1,2,3,4,5,....................k. no student gave more than k wrong answers. the total number of wrong answers given is__________
(A) a1+a2+a3+.......ak
(B) a1+a2+a3+........ak-1
(C) a1+a2+a3+.........ak+1
(D) none of these
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3 Answers
341
341I am not sure if you read the hidden portion. Anyway, here's the approach:
Consider the ak students who gave wrong answers to at least k questions. Since they could not have got more than k questions wrong (as per the 2nd sentence of the problem), this means ak students gave exactly k wrong answers contributing kak to the tally of wrong answers.
Now ak-1 students gave at least k-1 wrong answers - this group will comprise of those who have exactly (k-1) wrongs and those who have exactly k wrongs. Its obvious therefore that the number of students with exactly (k-1) wrongs is (ak-1-ak) so now we have a contribution of (k-1)(ak-1-ak).
Add 'em up and you will get a1+a2+...+ak