Permutations and Combinations-Quiz-17

Permutations and Combinations Quiz-17

Dear Readers,

Permutations and Combinations is one of the most important chapters of Algebra in the JEE syllabus and other engineering exams. For JEE Mains, it has 4% weightage and for JEE Advanced, it has 5% weightage..

Linked Comprehension Type This section contain(s) 9 paragraph(s) and based upon each paragraph, multiple choice questions have to be answered. Each question has atleast 4 choices (a), (b), (c) and (d) out of which ONLY ONE is correct.

Paragraph for Question No. 1

Q1. The number of zeros at the 100! is

•  22
•  23
•  24
•  26

Solution

Paragraph for Question No. 2

Q2.The number of permutations of the letters of the word INDIA taken three at a time must be

•  27
•  30
•  33
•  57

Solution

Paragraph for Question No. 3
Different words are beging formed by arranging the letters of the word “SUCCESS”. All the words obtained by written in the form of a dictionary
Q3.  The number of words in which the two C are together but no two 𝑆 are together, is

•  120
•  96
•  24
•  420

Solution

Paragraph for Question No. 4
We have to choose 11 players for cricket team from eight batsmen, six bowlers, four all rounder and two wicket keepers in the following conditions

Q4. The number of selections when almost one all rounder and one wicket keeper will play

•   ⁴𝐶₁ × ¹⁴𝐶₁₀ + ²𝐶₁ × ¹⁴𝐶₁₀ + ⁴𝐶₁ × ²𝐶₁ × ¹⁴𝐶₉ + ¹⁴𝐶₁₁
•   ⁴𝐶₁ × ¹⁵𝐶₁₁ + ¹⁵𝐶₁₁
•   ⁴𝐶₁ × ¹⁵𝐶₁₀ + ¹⁵𝐶₁₁
•  None of these

Solution

Paragraph for Question Nos. 5
Twelve 12 persons are to be arranged around two round tables such that one table can accommodate seven persons and another five persons only. Answer the following questions
Q5. Number of ways in which these 12 persons can be arranged is

•   ¹²𝐶₅6!4!
•  6!4!
•  ¹²C₅6!4!
•  None of these

Solution

Paragraph for Question Nos. 6
Five balls are to be placed in three boxes. Each box should hold all the five balls so that no box remains empty
Q6.  Number of ways if balls are different but boxes are identical is

•  30
•  25
•  21
•  35

Solution

Paragraph for Question Nos. 7 to 9
Let 𝑓(𝑛) denote the number of different ways in which the positive integer 𝑛 can be expressed as the sum of 1s and 2s. For example, 𝑓(4) = 5, since 4 = 2 + 2 = 2 + 1 + 1 = 1 + 2 + 1 = 1 + 1 + 2 = 1 + 1 + 1 + 1. Note that order of 1s and 2s is important Q7.𝑓:𝑁→𝑁 is

•   One-one and onto
•   One-one and into
•  Many-one and onto
•   Many-one and into

Solution

Q8.The value of 𝑓(𝑓(6)) is

•  400
•  350
•  377
•  None of these

Solution

Q9.The value of 𝑓(6) is

•  12
•  13
•  18
•  21

Solution

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