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Q.1. In a certain test, a_{i} students gave wrong answers to atleast i questions, where i = 1, 2, …, k. No student gave more than k wrong answers. The total number of wrong answers given is ................. (1982  2 Marks)
Ans. Sol. Number of students who gave wrong answers to exactly one question = a_{1} – a_{2}, Two questions = a_{2} – a_{3}
Three questions = a_{3}– a_{4}, k–1 question = a_{k–1}–a_{k} , k question = a_{k}
∴ Total number of wrong answers
= 1 (a_{1}– a_{2}) + 2 (a_{2} – a_{3}) + 3 (a_{3} – a_{4}) + .... (k – 1) (a_{k–1}– a_{k}) + k a_{k}
= a_{1 }+ a_{2 }+ a_{3} + ....a_{k}
Q.2. The side AB, BC and CA of a triangle ABC have 3, 4 and 5 interior points respectively on them. The number of triangles that can be constructed using these interior points as vertices is ................. (1984  2 Marks)
Ans. 205
Sol. We have total 3 + 4 + 5 = 12 points out of which 3 fall on one line, 4 on other line and 5 on still other line. So number of D‘s that can be formed using 12 such points are = ^{12}C_{3} – ^{3}C_{3 }– ^{4}C_{3} – ^{5}C_{3}
= 220 – 15 = 205
Q.3. Total number of ways in which six ‘+’ and four ‘–’ signs can be arranged in a line such that no two '–' signs occur together is ................. (1988  2 Marks)
Ans. 35
Sol. ‘+’ signs can be put in a row in 1 way, creating 7 ticked places to keep ‘–’ sign so that no two ‘–’ signs occur together
√ +√ + √ +√ +√ +√ +√
Out of these 7 places 4 can be chosen in ^{7}C_{4} ways.
∴ Required no. of arrangements are
= ^{7}C_{4} = ^{7}C_{3} =
Q.4. There are four balls of different colours and four boxes of colours, same as those of the balls. The number of ways in which the balls, one each in a box, could be placed such that a ball does not go to a box of its own colour is _________.
Ans. 9
Sol. KEY CONCEPT : We know that number of dearrangements of n objects
∴ No. of ways of putting all the 4 balls into boxes of different colour
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