MENU

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 thousand’splace 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 2C1

In the second box, thenumbers can be any of the four digits left, so the possibility is 4C1

In the third box, thenumbers can be any of the three digits left, so the possibility is 3C1

In the fourth box, thenumbers can be any of the two digits left, so the possibility is 2C1

Hence total number ofpossible outcomes is 2C1 × 4C1 × 3C1 × 2C1 =2 × 4 × 3 × 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 6C1

In the second seat,any one of five members can be seated, so the total number of possibilitiesis 5C1 ways.

In the third seat, anyone of four members can be seated, so the total number of possibilitiesis 4C1 ways.

In the fourth seat,any one of three members can be seated, so the total number of possibilitiesis 3C1 ways.

In the fifth seat, anyone of two members can be seated, so the total number of possibilities is 2C1 ways.

In the sixth seat,only one remaining person can be seated, so the total number of possibilitiesis 1C1 ways.

Hence, the totalnumber of possible outcomes = 6C1 × 5C1 × 4C1 × 3C1 × 2C1 × 1C1 =6 × 5 × 4 × 3 × 2 × 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 9C1=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 box can be filled with one of the eight available digits, so thepossibility is 8C1

The fourth digitcan be filled with one of the seven available digits, so the possibilityis 7C1

The fifth digitcan be filled with one of the six available digits, so the possibility is 6C1

The sixth digitcan be filled with one of the six available digits, so the possibility is 5C1

The seventh digitcan be filled with one of the six available digits, so the possibility is 4C1

The eighth digitcan be filled with one of the six available digits, so the possibility is 3C1

The ninth digitcan be filled with one of the six available digits, so the possibility is 2 C1

Hence the number oftotal possible outcomes is 9C1 × 9C1 × 8C1 × 7C1 × 6C1 =9 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 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 3C1 =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 × 2 = 6 such 2-digit numbers

Case 3: three-digitodd number

Ignore zero at one’splace 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 *3 * 2 = 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, one’s 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 * 3 * 2 = 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 –  6 = 12.

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.

Free - Previous Years Question Papers
Any questions? Ask us!
×