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­Why doesnt this approach work?

Use either equation (for non mutually exclusive events)

=> Total # of integers = # of integers that dont have 2 + # of integers that dont have 5 - # of integers that dont have 2 and 5

Since the problem is asking for # of integers that dont have EITHER a 2 or 5, wont this also include integers that dont have a 2 but have a 5 and integers that dont have a 5 but have a 2?

What logical mistake am I making?

I appreciate any response!!!
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Account9855
­Why doesnt this approach work?

Use either equation (for non mutually exclusive events)

=> Total # of integers = # of integers that dont have 2 + # of integers that dont have 5 - # of integers that dont have 2 and 5

Since the problem is asking for # of integers that dont have EITHER a 2 or 5, wont this also include integers that dont have a 2 but have a 5 and integers that dont have a 5 but have a 2?

What logical mistake am I making?

I appreciate any response!!!
­The phrase "do NOT contain either the digit 2 or the digit 5" means we are looking for numbers that do not contain either of these digits.­
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Is this the same as asking for numbers that contain neither 2 nor 5?
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Is this the same as asking for numbers that contain neither 2 nor 5?
Yes, numbers that do NOT contain either the digit 2 or the digit 5 is the same as numbers that contain neither 2 nor 5
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Hundreds digit option = 10-3=7 (cannot be 0 or 2 or 5)
Tens digit option = 10-2=8 (cannot be 2 or 5)
Units digit option = 10-2=8 (cannot be 2 or 5)
No constraint on repetition, so the number of integers = 7*8*8 = 448
soumyap09
How many of the integers from 100 to 999, inclusive, do NOT contain either the digit 2 or the digit 5?

A. 343
B. 448
C. 512
D. 648
E. 700
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soumyap09
How many of the integers from 100 to 999, inclusive, do NOT contain either the digit 2 or the digit 5?

A. 343
B. 448
C. 512
D. 648
E. 700

Responding to a pm:

The easiest method here is to use combinations only as discussed above too. I have discussed this method here: https://youtu.be/_C-kTA44OxE

We need all 3 digit numbers without 2 and 5.

__ ___ ___

The first digit can be chosen in 7 ways (1 to 9 except 2 and 5)
The second digit can be chosen in 8 ways (0 to 9 except 2 and 5)
The third digit can be chosen in 8 ways (0 to 9 except 2 and 5)

Total = 7 * 8 * 8
The answer must end with 8 and should be less than 8*8*10 = 640. Hence it must be 448

Answer (B)
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Bunuel
pls help me with the alternate approach
total - digits that have 2 or 5 in them
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Bunuel
pls help me with the alternate approach
total - digits that have 2 or 5 in them

Here is that approach: https://gmatclub.com/forum/how-many-of- ... l#p3395130

Hope it helps.
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pls help me with the alternate approach
total - digits that have 2 or 5 in them
Total = numbers between 100 and 999 inclusive = 900.

200-299 and 500-599 will each have 100 numbers which contain 2 or 5.

For remaining each set of 100 count numbers having 2 or 5.
Let's look at 100 - 199, for instance.

Start with 2s:
102, 112, 122, 132, 142, 152, 162, 172, 182, 192 = 10
120 - 129 has 10 numbers but we have already counted 122 above, so for this range we will only count 9.
Total unique numbers with 2s in 100-199 = 19.

For 5s: Just like above we will get 19.
But in these 19, we have 125 and 152 - which are already counted in 2s above, and will be double counted in 5s again.

So unique numbers in 100-199 having 2 or 5 = 19 + 19 - 2 = 36.

We have 100-199, 300-399, 400-499, 600-699, 700-799, 800-899, 900-999 = 7 such sets of 100 numbers.
So total numbers having 2 or 5 = (2 x 100) + (36 x 7) = 200 + 252 = 452

Numbers not having 2 or 5 = 900 - 452 = 448
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