Math, asked by tejdeep23, 2 months ago

the numbers of all possible natural numbers using digits 9,8 and 0 when repetition of digits is not allowed as​

Answers

Answered by BhoomiGupta01
2

Answer:

The given digits are 7,0,6 and repetition of digits is not allowed.

The one-digit numbers that can be formed are 7 and 6.

We are required to write 2−digit numbers.

Out of the given digits, the possible ways of choosing the two digits are

7,0;6,0;6,7

Using the digits 7 and 0, the numbers are 70.

Similarly, Using the digits 6 and 7, the numbers are 67 and 76.

Hence, all possible 2−digit numbers are

60,70,67,76

Now, we are required to write 3−digit numbers using the digits 7,0,6 and the repetition of the digits is not allowed. Keeping 0 at unit's place, 3−digit number obtained are 670 and 760.

Keeping 6 at unit's place, 3−digit number obtained are 706.

Keeping 7 at unit's place, 3−digit number obtained are 607.

Hence ,all possible 3−digit numbers are: 670,764,706 and 607.

All possible numbers using the digits 7,0 and 6 are

6,7,76,67,70,60,706,607,760,670. I hope you help you

Step-by-step explanation:

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Answered by appupatil68
1

Answer:

Answer is 4

Reason-9,0,8

=980,908

=890,809

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