The total number of 3 digit numbers which have two or more consecutive digits identical is:

a)
171

b)
170

c)
90

d)
180

e)
None of these

Correct answer is option 'A'. Can you explain this answer?

Question

The total number of 3 digit numbers which have two or more consecutive digits identical is:a)171b)170c)90d)180e)None of theseCorrect answer is option 'A'. Can you explain this answer?

Answer

Can we do this method too, I just tried it out,

Front to Back, I mean tens to units place common for 0-9 is , 9 because there are hundreds place can't be zero

This patterns happens tens times, so 9(10)= 90

Back to Front,

I mean tens to units place common for 0-9 is , 9 because there are hundreds place can't be zero

This patterns happens tens times, so 9(10)= 90 again

So 2 times 90 is 180 and there is an intersection which is observed, 111, 222, 333... which I call common Front to Back or common Back to Front, and this pattern happens 9 times, so subtract that and we get the answer as 180-9= 171, EduRev please as your team to verify my answer.

Answer

In each set of 100 numbers, there are 10 numbers whose tens digit and unit digit are the same. Again, in the same set, there are 10 numbers whose hundreds and tens digits are the same.

But one number in each set of 100 numbers whose Hundreds, Tens and Unit digit are the same as 111, 222, 333, 444, etc.
Hence, there are exactly (10 + 10 - 1) = 19 numbers in each set of 100 numbers. Further, there are 9 such sets of numbers.
Therefore such total numbers = 19 × 9 = 171

Alternatively,
9 × 10 × 10 - 9 × 9 × 9 = 900 - 729 = 171

Answer

Solution:

To find the desired number of 3 digit numbers which have two or more consecutive digits identical, we can use the following approach:

- Case 1: The first two digits are identical.
In this case, the first and second digits can be chosen in 9 ways (excluding 0), and the third digit can be chosen in 10 ways (including 0). Therefore, the total number of such 3 digit numbers is 9 × 10 = 90.

- Case 2: The last two digits are identical.
In this case, the second and third digits can be chosen in 9 ways (excluding 0), and the first digit can be chosen in 10 ways (including 0). Therefore, the total number of such 3 digit numbers is 9 × 10 = 90.

- Case 3: The second and third digits are identical.
In this case, the first digit can be chosen in 9 ways (excluding 0), and the second and third digits can be chosen in 10 ways (including 0). Therefore, the total number of such 3 digit numbers is 9 × 10 = 90.

However, we have counted some numbers twice in the above cases. Specifically, we have counted the numbers with all three digits identical (e.g. 111) three times, and we have counted the numbers with the first and last digits identical (e.g. 121) twice. Therefore, we need to subtract these numbers from the total count.

- Numbers with all three digits identical: There is only 1 such number, namely 111.

- Numbers with the first and last digits identical: The first and last digits can be chosen in 9 ways (excluding 0), and the middle digit can be chosen in 10 ways (including 0). Therefore, the total number of such 3 digit numbers is 9 × 10 = 90.

Hence, the total number of 3 digit numbers which have two or more consecutive digits identical is:

90 + 90 + 90 - 1 - 90 = 171

Therefore, the correct answer is (a) 171.

Answer

ये क्या होता हे इसे क्या बोलते हे

Answer

Because in every case you will find first two digit same that's why and is 171

and its consecutive no are 172 173...

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