Permutations and Combinations Class 11 OP Malhotra Exe-12A ISC Maths Solutions Ch-12 Solutions. In this article you would learn about Fundamental Principle and its Counting. Step by step solutions of latest textbook has been given as latest syllabus. Visit official Website CISCE for detail information about ISC Board Class-11 Mathematics.
Permutations and Combinations Class 11 OP Malhotra Exe-12A ISC Maths Solutions Ch-12
| Board | ISC |
| Publications | S Chand |
| Subject | Maths |
| Class | 11th |
| Chapter-12 | Permutations and Combinations |
| Writer | O.P. Malhotra |
| Exe-12(A) | Fundamental Principle and its Counting. |
Exercise- 12A
Permutations and Combinations Class 11 OP Malhotra Exe-12A Solution.
Que-1: Two persons go in a railway carriage where there are 6 vacant seats. In how many different ways can they seat themselves ?
Sol: Clearly first person can sit in 6 ways and second person can sit in 5 ways
∴ required no. of ways = 6 × 5 = 30
Que-2: In how many ways can 2 prizes be awarded to 9 contestants provided no contestant gets both the prizes ?
Sol: Since first prize can be given to any of the 9 contestants.
since no contestant gets both the prizes thus second prize can be given to remaining 8 contestants
∴ required no. of ways = 9 × 8 = 72
Que-3: There are three mathematics teachers in a college in which there are 6 classes. In how many different ways can they choose the classes, provided one teaches one class only ?
Sol: Since first mathematics teacher can take any of the 6 classes so this can be done in 6 ways.
For second maths teacher, we have remaining 5 classes
since it is given that one teaches one class only and third maths teacher can take any one of four classes
∴ required no. of ways = 6 × 5 × 4 = 120
Que-4: How many words (with or without meaning) of three distinct letters of the English alphabets are there ?
Sol: There are 26 English alphabets
∴ required no. of words of three distinct letters = 26 × 25 × 24 = 15600
Que-5: How many numbers are there between 100 and 1000 such that 7 is in the units place ?
Sol: Clearly the numbers between 100 and 1000 are 3 digit numbers.
it is given that 7 in the unit place
Thus tens place can be filled by any of the 10 digits and this can be done in 10 ways.
Hundred place can be filled by 9 ways. (Excluding 0 otherwise number becomes 2-digit)
∴ Required no. of numbers = 9 × 10 × 1 = 90
Que-6: How many integers of four digits each can be formed with the digits 0,1, 3, 5, 6 (assuming no repetitions) ?
Sol: Given digits are 0, 1, 3, 5, 6
Thousand place can be filled by 4 ways (excluding 0)
hundred place can be filled by 4 ways (including 0)
since repetition is not allowed.
Ten place can be filled by 3 ways and unit place can be filled by 2 ways,
∴ required no. of numbers = 4 × 4 × 3 × 2 = 96
Que-7: How many automobile license plates can be made if the inscription on each contains two different letters followed by three different digits ?
Sol: Since inscription are to be made out of 26 different letters and 10 different digits. Since digit 0 can’t be at hundred place.
∴ required no. of license plates can be made = 26 × 25 × 9 × 9 × 8 = 4,21,200
Que-8: Find the number of ways of arranging 6 players to throw the cricket ball so that the oldest player may not throw first.
Sol: For first place, we have five players (excluding oldest)
For the remaining 5 places we have five players (including the oldest)
∴ required no. of ways = 5 × 5 × 4 × 3 × 2 × 1 = 600
Que-9: How many three-digit numbers can be formed without using the digits 0, 2, 3, 4, 5 and 6 ?
Sol: The remaining available digits are 1, 7, 8, 9 unit place can be filled by 4 ways
Tens and hundred place can be filled by 4 ways each.
Thus required no. of three digit numbers = 4 × 4 × 4 = 64
Que-10: Find the number of even positive integers which have three digits.
Sol: We want to find 3 digit even numbers.
Clearly unit place can be filled by digits 0, 2, 4, 6, 8 and this can be done by 5 ways.
Hundred place can be filled by 9 ways (excluding 0) and tens place can be filled 10 ways (including 0)
∴ required no. of even integers = 9 × 10 × 5 = 450
Que-11: How many 2-digit numbers can be formed from the digits 8,1, 3, 5 and 4 assuming
(a) repetition of digits is allowed ?
(b) repetition of digits is not allowed ?
Sol: (a) Given digits are 8, 1, 3, 5 and 4 and repetition of digits is allowed
Thus unit place can be filled by 5 ways
Tens place can be filled by 5 ways.
∴ required no. of 2-digit numbers = 5 × 5 = 25
(b) Repetition of digits is not allowed
unit place can be filled by 5 ways and tens place can be filled by 4 ways.
∴ required no, of such 2-digit numbers = 4 × 5 = 20
Que-12: There are 12 true-false questions in an examination. How many sequences of answers are possible ?
Sol: For each true-false question, we have 2 ways to answer it either true or false.
We have given 12 true-false questions.
∴ required no. of possible sequence of answers = 212 = 4096
Que-13: How many four-digit even integers can be formed using the digits 0, 1, 2, 3, 4, 5 ?
Sol: Given digits are 0, 1, 2, 3, 4, 5
We want to find four digit even integers so its unit place contain only 0, 2 or 4
Thus unit place can be filled by 3 ways
Thousand place can be filled by 5 ways (excluding 0) and tens
place can be filled by 6 ways.
Tens place can be filled by 6 ways.
∴ required no. of four digit even integers = 5 × 6 × 6 × 3 = 540
Que-14: To pass an examination a student has to pass in each of the 3 papers. In how many ways can a student fail in the examination ?
Sol: For each paper, students have 2 choice either pass or fails.
Thus total no. of ways in which a student attempt all the three papers = 2 × 2 × 2 = 23
It is given that to pass an examination a student has to pass in each of the three papers and this
can be done in only one ways i.e. (PPP)
Thus required no. of ways in which a student fails in the examination = 23 – 1 = 7
Que-15: How many seven-digit phone numbers are possible if 0 and 1 cannot be used as the first digit and the first three digits cannot be 555, 411, or 936 ?
Sol: For first three digits :
Unit place can be filled by 8 ways as 0 and 1 cannot be at unit place.
Tens place and hundred place can be filled by 10 ways each. Further it is given that first three digits cannot be 555, 411 or 936
∴ number of ways in which first three digits can be arranged = 10 × 10 × 8 – 3 = 797
Further last four digit can be filled by 10 ways each.
Thus, required no. of such 7-digit numbers = 10 × 10 × 10 × 10 × 797 = 797,0000
Que-16: There are five routes for a journey from station A to station B, In how many different ways can a man go from A to B and return, if for returning
(i) any of the routes is taken
(ii) the same route is taken
(iii) the same route is not taken ?
Sol: (i) Total no. of ways in which a man can go from A to B = 5
since for returning, any of the route is taken
∴ Total no. of ways in which a man can go from B to A = 5
Hence required no. of ways = 5 × 5 = 25
(ii) For returning from B to A, same route is taken by man and this can be done in only 1 way. Thus required no. of such ways = 5 × 1 = 5
(iii) For returning from B to A, same route is not taken by man and this can be done in (5 – 1) = 4 ways.
Hence, required no. of ways = 5 × 4 = 20
Que-17: How many 9-digit numbers of different digits can be formed ?
Sol: Here 9th place can be filled by 9 ways (excluding 0)
8th place can be filled by 9 ways (including 0)
7th place can be filled by 8 ways
6th place can be filled by 7 ways
5th place can be filled by 6 ways
4th place can be filled by 5 ways
3th place can be filled by 4 ways
2th place can be filled by 3 ways
unit place can be filled by 2 ways
Thus required no. of 9 digit numbers = 9 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 = 3265920
Que-18: For a set of five true-or false-questions, no student has written the all-correct answers, and no two students have given the same sequence of answers. What is the maximum number of students in the class, for this to be possible ?
Sol: For each true-false question every student have 2 choices to answer a question, either true or false.
we have 5 true-false questions and there is only one sequence of all correct answers.
∴ Required no. of such possible ways = 25 – 1 = 32 – 1 = 31
–: End of Permutation and Combinations Class 11 OP Malhotra Exe-12A ISC Maths Ch-12 Solutions. :–
Return to :- OP Malhotra ISC Class-11 S Chand Publication Maths Solutions
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