In this page, you can find the complete solutions of the first exercise of Permutation and Combination chapter from Basic Mathematics Grade XII book published and distributed by Sukunda Pustak Bhawan.
In the above-mentioned book, permutation and combination is the first chapter and has three exercises only. Out of which, this is the solution to the first exercise.
Disclaimer:
Exercise 1.1
1. A football stadium has four entrance gates and nine exits. In how many different ways a man can enter and leave the stadium?
Answer: A man can enter the stadium from any one of the four entrance gates (m = 4) and can leave the stadium from any one of the nine exit gates (n = 9). By the basic principle of counting, a man can enter and leave the stadium in mxn = 4x9 = 36 different ways.
2. There are six doors in a hostel. In how many ways can a student enter the hostel and leave by a different door?
Answer: A student can enter the hostel by any one of the six doors (m = 6). To exit the hostel from a different door, he cannot exit from the same door he entered. So, the total choices for exit doors are five (n = 5). By the basic principle of counting, a student can enter and leave the hostel by a different door in mxn = 6x5 = 30 different ways.
3. In how many ways can a man send three of his children to seven different colleges in a certain town?
Answer: A man has seven different choices for his first child (m = 7). To send his second child to a different college, he cannot send him to the same college where he sent his first child. So, the second child can go into any one of the six remaining colleges (n = 6). Similarly, the third child can choose from the remaining five colleges (o = 5). By the basic principle of counting, a man can send three of his children to seven different colleges in a certain town in mxnxo = 7x6x5 = 210 ways.
4. Suppose there are five main roads between cities A and B. In how many ways can a man go from one city to the other and return by a different road?
Answer: A man can from city A to city B via any one of the five main roads (m = 5). During his return, if he wishes to come back through a different road then he has to choose from the remaining four main roads (n = 4). By the basic principle of counting, a man can go from one city to the other and return by a different road in mxn = 5x4 = 20 ways.
5. There are five main roads between cities A and B and 4 between B and C. In how many ways can a person drive from A to C and return without driving on the same road twice?
Answer: Let us consider the man goes from A-B-C and returns from C-B-A. When going from A to B, he can choose one of the five main roads (m = 5). When going from B to C, he can choose one of the four main roads (n = 4). During his return, he will have one less road available if he does not drive on the same road twice. So, from C to B, he has three choices (o = 3). Similarly, from B to A, he has four choices (p = 4). By the basic principle of counting, a man can drive from A to C and return without driving on the same road twice in mxnxoxp = 5x4x3x4 = 240 ways.
6. How many numbers of at least three different digits can be formed from the integer 1,2,3,4,5,6?
Answer: In a number containing different digits, no digits should be repeated. The least number of digits in a number can be 6 and the maximum number of digits can be 6 (without repetition).
To construct a number of three different digits, the ones place can be taken by any of the six digits (m = 6), the tens place can be taken by the remaining five digits (n = 5) and the hundreds place can be taken by the remaining four digits (o = 4). By the basic principle of counting, the number of three different digits formed from integer 1,2,3,4,5,6 is mxnxo = 6x5x4 = 120.
Similarly, the number of four, five, and six different digits formed from integers 1,2,3,4,5,6 is 360, 720, and 720, respectively. Hence, the total number of numbers of at least three different digits that can be formed from the integer 1,2,3,4,5,6 is 120 + 360 + 720 + 720 = 1920.
7. How many numbers of three digits less than 500 can be formed from integer 1,2,3,4,5,6?
Answer: A number of three digits less than 500 is anywhere between 99 and 500. The hundreds place can only be occupied by four of the six integers to make it less than 500 (m = 4). The tenth place can be filled by any of the remaining five integers (n = 5). Similarly, the ones place can be filled by four integers (o = 4). By the basic principle of counting, the number of three digits less than 500 that can be formed from the integer 1,2,3,4,5,6 is mxnxo = 4x5x4 = 80.
8. Of the numbers formed by using all the figures 1,2,3,4,5 only once, how many are even?
Answer: Using all five figures, we can make numbers of 5 different digits. A number is said to be even if its ones place contains an even digit. In a five-digit number, the ones place can be occupied by 2 or 4 (m = 2) and the tens place can be occupied by the remaining four integers (n = 4). Similarly, the hundreds, thousands, and ten-thousands places can be occupied by 3,2, and 1 digit respectively. By the basic principle of counting, the number of five-digit even number is mxnx3x2x1 = 2x4x3x2x1 = 48.
Question 9 and 10 have been left for you to practice yourself.
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