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Fill in the Blanks: Permutations and Combinations | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE

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Q.1. In a certain test, ai 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 = a1 – a2, Two questions = a2 – a3
Three questions = a3– a4, k–1 question = ak–1–ak , k question = ak
∴ Total number of wrong answers
=  1  (a1– a2) + 2 (a2 – a3) + 3 (a3 – a4) + .... (k – 1) (ak–1– ak) + k ak
= a+ a+ a3 + ....ak

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 =   12C33C4C35C3

= 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 7C4 ways.
∴ Required no. of arrangements are
= 7C4 = 7C3 =

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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