Math, asked by shaikirshad7, 1 year ago

Find the number of even positive numbers which have five digits.​

Answers

Answered by bsola6178
8

Answer: 45000

Step-by-step explanation: A number is said to be an even number only when the last digit is either 0 or 2 or 4 or 6 or 8.

Case 1:

Count of Even numbers ending with 0 ….

1st digit can be filled in 9 ways..Either by 1 or 2 or 3 or 4 or 5 or 6 or 7 or 8 or 9.

2nd digit can be filled in 10 ways ..Either by 0 or 1 or 2 or 3 or 4 or 5 or 6 or 7 or 8 or 9.

3rd digit can be filled in 10 ways ..Either by 0 or 1 or 2 or 3 or 4 or 5 or 6 or 7 or 8 or 9.

4rth digit can be filled in 10ways either by 0 or 1 or 2 or 3 or 4 or 5 or 6 or 7 or 8 or 9

Last digit i.e 5th digit is fixed for 0.

So Case 1 has 9*10*10*10*1 = 9000 numbers..

Case 2:

Count of Even numbers ending with 2….

1st digit can be filled in 9 ways..Either by 1 or 2 or 3 or 4 or 5 or 6 or 7 or 8 or 9.

2nd digit can be filled in 10 ways ..Either by 0 or 1 or 2 or 3 or 4 or 5 or 6 or 7 or 8 or 9.

3rd digit can be filled in 10 ways ..Either by 0 or 1 or 2 or 3 or 4 or 5 or 6 or 7 or 8 or 9.

4rth digit can be filled in 10ways either by 0 or 1 or 2 or 3 or 4 or 5 or 6 or 7 or 8 or 9

Last digit i.e 5th digit is fixed for 2.

So Case 2 has 9*10*10*10*1 = 9000 numbers..

Case 3:

Count of Even numbers ending with 4 ….

1st digit can be filled in 9 ways..Either by 1 or 2 or 3 or 4 or 5 or 6 or 7 or 8 or 9.

2nd digit can be filled in 10 ways ..Either by 0 or 1 or 2 or 3 or 4 or 5 or 6 or 7 or 8 or 9.

3rd digit can be filled in 10 ways ..Either by 0 or 1 or 2 or 3 or 4 or 5 or 6 or 7 or 8 or 9.

4rth digit can be filled in 10ways either by 0 or 1 or 2 or 3 or 4 or 5 or 6 or 7 or 8 or 9

Last digit i.e 5th digit is fixed for 4.

So Case 3 has 9*10*10*10*1 = 9000 numbers..

Case 4:

Count of Even numbers ending with 6….

1st digit can be filled in 9 ways..Either by 1 or 2 or 3 or 4 or 5 or 6 or 7 or 8 or 9.

2nd digit can be filled in 10 ways ..Either by 0 or 1 or 2 or 3 or 4 or 5 or 6 or 7 or 8 or 9.

3rd digit can be filled in 10 ways ..Either by 0 or 1 or 2 or 3 or 4 or 5 or 6 or 7 or 8 or 9.

4rth digit can be filled in 10ways either by 0 or 1 or 2 or 3 or 4 or 5 or 6 or 7 or 8 or 9

Last digit i.e 5th digit is fixed for 6.

So Case 4 has 9*10*10*10*1 = 9000 numbers..

Case 5:

Count of Even numbers ending with 8….

1st digit can be filled in 9 ways..Either by 1 or 2 or 3 or 4 or 5 or 6 or 7 or 8 or 9.

2nd digit can be filled in 10 ways ..Either by 0 or 1 or 2 or 3 or 4 or 5 or 6 or 7 or 8 or 9.

3rd digit can be filled in 10 ways ..Either by 0 or 1 or 2 or 3 or 4 or 5 or 6 or 7 or 8 or 9.

4rth digit can be filled in 10ways either by 0 or 1 or 2 or 3 or 4 or 5 or 6 or 7 or 8 or 9

Last digit i.e 5th digit is fixed for 8.

So Case 5 has 9*10*10*10*1 = 9000 numbers..

Now,the total number of 5 digit even numbers will be summation of all the possible ways…i.e case 1 + case 2 + case 3 + case 4 + case 5 = 9000+9000+9000+9000+9000= 45000 numbers possible

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Answered by thatsgirijag
2

Answer:

9*10*10*10*5

Step-by-step explanation:

number is even if the digit at its unit place is 1 out of 02468

number of ways of filling unit place is 5

number of ways of filling tense please is 10

number of ways of killing hundreds place is 10

number of ways of filling thousands please is 10

number of ways of filling 10,000 place is 9 (zero cannot be put at 10,000 place)

by the fundamental principle of multiplication total number of ways is 9*10*10*10*5

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