How many 1 are there while we count from 1 to 100????
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There are two cases.
When one is at first place. Here…As we have to consider two digit numbers so 1 at first place and any from the 10 digits (0–9) can come at second place so in this way we got 10 possible permutations.
When 1 is at second place. Here at first place we will have 9 digits(1–9). Hence it will give 9 permutations. From which 11 is common in both. So now the total permutations are
10 + 9 - 1 +1 = 19 (+1, because 11 has 1, 2 times)
Now 1 case is 100 and one is 1 itself. So finally we have 20 permutations.
So 1 comes 21 times between 1 and 100. (If 1 and 100 both are inclusive).
Else we have 19 times.
When one is at first place. Here…As we have to consider two digit numbers so 1 at first place and any from the 10 digits (0–9) can come at second place so in this way we got 10 possible permutations.
When 1 is at second place. Here at first place we will have 9 digits(1–9). Hence it will give 9 permutations. From which 11 is common in both. So now the total permutations are
10 + 9 - 1 +1 = 19 (+1, because 11 has 1, 2 times)
Now 1 case is 100 and one is 1 itself. So finally we have 20 permutations.
So 1 comes 21 times between 1 and 100. (If 1 and 100 both are inclusive).
Else we have 19 times.
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