Class XI Chapter 7 – Permutation and Combination Maths ______________________________________________________________________________ Question 3: How many 3-digit even numbers can be made using the digits 1, 2, 3, 4, 6, 7, if no digit is repeated? Solution 3: 3-digit even numbers are to be formed using the given six digits, 1, 2, 3, 4, 6, and 7, without repeating the digits. Then, units can be filled in 3 ways by any of the digits, 2, 4 or 6. Since the digits cannot be repeated in the 3-digit numbers and units place is already occupied with a digit (which is even), the hundreds and tens place is to be filled by the remaining 5 digits. Therefore, the number of ways in which hundreds and tens place can be filled with the remaining 5 digits is the permutation of 5 different digits taken 2 at a time. Number of ways of filling hundreds and tens place 5 P2 5! 5! 5 2 ! 3! 543! 20 3! Thus, by multiplication principle, the required number of 3-digit numbers is 3 20 60 . Question 4: Find the number of 4-digit numbers that can be formed using the digits 1, 2, 3, 4, 5 if no digit is repeated. How many of these will be even? Solution 4: 4-digit numbers are to be formed using the digits, 1, 2, 3, 4 and 5. There will be as many 4-digit numbers as there are permutations of 5 different digits taken 4 at a time. Therefore, required number of 4 digit numbers 5 P4 5! 5! 5 4 ! 1! 12345 120 Among the 4-digit numbers formed by using the digits, 1, 2, 3, 4, 5 even numbers end with either 2 or 4. The number of ways in which units place is filled with digits is 2. Since the digits are not repeated and the units place is already occupied with a digit (which is even), the remaining places are to be filled by the remaining 4 digits. Therefore, the number of ways in which the remaining places can be filled is the permutation of 4 different digits taken 3 at time. Number of ways of filling the remaining places 4321 24 4 P3 4! 4! 4 3 ! 1! Thus, by multiplication principle, the required number of even numbers is 24 2 48 . Printed from Vedantu.com. Register now to book a Free LIVE Online trial session with a Top tutor.

Class XI Chapter 7 – Permutation and Combination Maths ______________________________________________________________________________ Question 5: From a committee of 8 persons, in how many ways can we choose a chairman and a vice chairman assuming one person cannot hold more than one position? Solution 5: From a committee of 8 persons, a chairman and a vice chairman are to be chosen in such as way that one person cannot hold more than one position. Here, the number of ways of choosing a chairman and a vice chairman is the permutation of 8 different objects taken 2 at a time. Thus, required number of ways 8 P2 8! 8! 876! 8 2 ! 6! 6! 8 7 56 Question 6: Find n if Solution 6: n1 n 3 4 P : P 1:9 n1 n 3 4 P : P 1:9 P3 P n1 n 4 n 1 9 n 1! 13 ! 1 n! 9 n 4! n1 ! n4 ! 1 n 4 ! n! 9 n n 1 1 n 9 n 9 1! 1 n1 ! 9 . Printed from Vedantu.com. Register now to book a Free LIVE Online trial session with a Top tutor.

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