RD Chapter 16 Permutations Ex 16.2 Solutions
Question - 21 : - How many four-digit numbers can be formed with the digits 3, 5, 7, 8, 9 which are greater than 8000, if repetition of digits is not allowed?
Answer - 21 : -
The required numbersare greater than 8000.
So, the thousandтАЩsplace can be filled with 2 digits: 8 or 9.
Let us assume fourboxes, now in the first box can either be one of the two numbers 8 or 9, sothere are two possibilities which is┬а2C1
In the second box, thenumbers can be any of the four digits left, so the possibility is┬а4C1
In the third box, thenumbers can be any of the three digits left, so the possibility is┬а3C1
In the fourth box, thenumbers can be any of the two digits left, so the possibility is┬а2C1
Hence total number ofpossible outcomes is┬а2C1┬а├Ч┬а4C1┬а├Ч┬а3C1┬а├Ч┬а2C1┬а=2 ├Ч 4 ├Ч 3 ├Ч 2 = 48.
Question - 22 : - In how many ways can six persons be seated in a row?
Answer - 22 : -
Let us assume thereare six seats.
In the first seat, anyone of six members can be seated, so the total number of possibilities is┬а6C1
In the second seat,any one of five members can be seated, so the total number of possibilitiesis┬а5C1┬аways.
In the third seat, anyone of four members can be seated, so the total number of possibilitiesis┬а4C1┬аways.
In the fourth seat,any one of three members can be seated, so the total number of possibilitiesis┬а3C1┬аways.
In the fifth seat, anyone of two members can be seated, so the total number of possibilities is┬а2C1┬аways.
In the sixth seat,only one remaining person can be seated, so the total number of possibilitiesis┬а1C1┬аways.
Hence, the totalnumber of possible outcomes =┬а6C1┬а├Ч┬а5C1┬а├Ч┬а4C1┬а├Ч┬а3C1┬а├Ч┬а2C1┬а├Ч┬а1C1┬а=6 ├Ч 5 ├Ч 4 ├Ч 3 ├Ч 2 ├Ч 1 = 720.
Question - 23 : - How many 9-digit numbers of different digits can be formed?
Answer - 23 : -
In a nine-digitnumber, 0 cannot appear in the first digit and the repetition of digits is notallowed. So, the number of ways of filling up the first digit is┬а9C1=9
Now, 9 digits are leftincluding 0. So, the second digit can be filled with any of the remaining 9digits in 9 ways.
Similarly, thethird┬аbox can be filled with one of the eight available digits, so thepossibility is┬а8C1
The fourth┬аdigitcan be filled with one of the seven available digits, so the possibilityis┬а7C1
The fifth┬аdigitcan be filled with one of the six available digits, so the possibility is┬а6C1
The sixth┬аdigitcan be filled with one of the six available digits, so the possibility is┬а5C1
The seventh┬аdigitcan be filled with one of the six available digits, so the possibility is┬а4C1
The eighth┬аdigitcan be filled with one of the six available digits, so the possibility is┬а3C1
The ninth┬аdigitcan be filled with one of the six available digits, so the possibility is┬а2┬аC1
Hence the number oftotal possible outcomes is┬а9C1┬а├Ч┬а9C1┬а├Ч┬а8C1┬а├Ч┬а7C1┬а├Ч┬а6C1┬а=9 ├Ч 9 ├Ч 8 ├Ч 7 ├Ч 6 ├Ч 5 ├Ч 4 ├Ч 3 ├Ч 2 = 9 (9!)
Question - 24 : - How many odd numbers less than 1000 can be formed by using the digits 0, 3, 5, 7 when repetition of digits is not allowed?
Answer - 24 : -
Any number less than1000 may be any of a number from one-digit number, two-digit number and three-digitnumber.
Case 1: One-digit oddnumber
In order to make thenumber odd, the last digit has to be either of (3, 5, 7)
In the first boxeither of the three digits (3,5,7) can be placed, so the possibility is┬а3C1┬а=3 possible ways.
Case 2: two-digit oddnumber
Tens place can befilled up by 3 ways (using any of the digit among 3, 5 and 7) and then the onesplace can be filled in any of the remaining 2 digits.
So, there are3┬а├Ч┬а2 = 6 such 2-digit numbers
Case 3: three-digitodd number
Ignore zero at oneтАЩsplace for some instance.
Hundreds place can befilled up in 3 ways (using any of any of the digit among 3, 5 and 7), then tensplace in 3 ways by using remaining 3 digits (after using a digit, there will bethree digits) and then the ones place in 2 ways.
So, there are a totalof 3┬а*3┬а*┬а2 = 18 numbers of 3-digit numbers which includes bothodd and even numbers (ones place digit are zero). In order to get the oddnumbers, it is required to ignore the even numbers i.e. numbers ending with zero.
To obtain the even3-digit numbers, oneтАЩs place can be filled up in 1 way (only 0 to be filled),hundreds place in 3 ways (using any of the digit among 3, 5, 7) and then tensplace in 2 ways (using remaining 2 digits after filling up hundreds place).
So, there are a totalof 1┬а*┬а3┬а*┬а2 = 6 even 3-digit numbers using the digits 0,3, 5 and 7 (repetition not allowed)
Then, number ofthree-digit odd numbers using the digits 0, 3, 5 and 7 (repetition not allowed)= 18 тАУ┬а 6 = 12.
тИ┤ The odd numbers lessthan 1000 that can be formed by using the digits 0, 3, 5, 7 when repetition ofdigits is not allowed are 3 + 6 + 12 = 21.