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Re: Of the three-digit integers greater than 750, how many have at least [#permalink]
chetan2u wrote:
Bunuel wrote:
Of the three-digit integers greater than 750, how many have at least two digits that are equal to each other?

(A) 56
(B) 70
(C) 72
(D) 74
(E) 78

Are You Up For the Challenge: 700 Level Questions


We will divide it into three parts 750-799, 800-899 and 900-999

Two ways almost similar to the ones explained above..

(I) Find the numbers with unique digits..


750-799 - 1st digit is 7, tens digit can be any of 5, 6, 8 or 9--so 4 ways, and the ones digit will be any of the digits except first and tens digit so 10-2=8 ways..
Numbers = 1*4*8=32 ways

Hi chetan,
in the text highlighted above in RED. if we take 1*4*8, don't we have to subtract 1 from this because this will consider 750 as well and in the question is asked for number greater than 750.?
I'll appreciate if you can clear my doubt.

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Re: Of the three-digit integers greater than 750, how many have at least [#permalink]
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Expert Reply
Bunuel wrote:
Of the three-digit integers greater than 750, how many have at least two digits that are equal to each other?

(A) 56
(B) 70
(C) 72
(D) 74
(E) 78

Are You Up For the Challenge: 700 Level Questions


Solution:

In the 700s, if the tens and ones digits are the same, we have: 755, 766, 777, 788, and 799, i.e., a total of 5 such numbers. If the hundreds and tens digits are the same, we have: 770, 771, …, 777, 778 and 779, i.e., a total of 10 such numbers. If the hundreds and ones digits are the same, we have: 757, 767, 777, 787 and 797, i.e., a total of 5 such numbers. It seems there are a total of 20 numbers in the 700s that satisfy the criterion; however, we can see that 777 has been counted three times, so we have to subtract 2 from 20. Thus, the actual number of numbers in the 700s is 18.

In the 800s, if the tens and ones digits are the same, we have: 800, 811, …, 888, and 899, i.e., a total of 10 such numbers. If the hundreds and tens digits are the same, we have 880, 881, …, 888, and 889, i.e., a total of 10 such numbers. If the hundreds and ones digits are the same, we have 808, 818, …, 888, and 898, i.e., a total of 10 such numbers. It seems there are a total of 30 numbers in the 800s that satisfy the criterion; however, we can see that 888 has been counted three times, so we have to subtract 2 from 30. Thus, the actual number of numbers in the 800s is 28.

Like the 800s, there should be another 28 numbers in the 900s. Therefore, there are a total of 18 + 28 + 28 = 74 such numbers.

Alternate Solution:

Notice that there are 999 - 751 + 1 = 249 three-digit integers that are greater than 750. If n is the number of three-digit integers that are greater than 750 such that no two digits of n are the same, then we must have:

The number of three-digit integers greater than 750 where at least two digits are the same = 249 - n

So, let’s calculate n.

In the 700s, the tens digit can be 5, 6, 8, or 9, and the units digit can be any one of the ten digits besides 7 and tens digit. Thus, there are 4 x 8 = 32 choices. However, one of these choices is 750 (which is not greater than 750); therefore there are 32 - 1 = 31 integers greater than 750 where no two digits is the same.

In the 800s, the tens digit can be any digit besides 8, and the units digit can be any digit besides 8 and the tens digit. Thus, there are 9 x 8 = 72 choices.

The above holds also for the 900s; thus, there are 72 integers between 900 and 999 where no two digits are the same as well.

Thus, n = 31 + 72 + 72 = 175. Hence, there are 249 - 175 = 74 three digit integers greater than 750 where at least two of the digits are the same.

Answer: D
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Of the three-digit integers greater than 750, how many have at least [#permalink]
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Soln:

We need three digit nos greater than 750 ,where atleast two digits are same

(so this includes all numbers where either two digits are same and all three digits are same)

Digits greater than 750 includes:
1) Numbers where all three digits are different
2) Numbers where two digits are same and one is different
3) All three digits same.

Lets first find - Total three digits numbers greater than 750

Last term- First Term+ 1

Last three digit term is 999, First term greater than 750 is 751

So 999- 751 +1=249

Total Terms from 751 to 999= 249

Now we cal 1) Numbers where all three digits are different

Between 751 to 799 :

First digit can be only one i.e 7
Second digit can be any of four - 5,6,8,9 ( but not 7 bcaz we want all three digits different)
Third digit can be any of remaining 8 ( from 0 to 9)

Total three different digit numbers between 7510 to 799 will be 1*4*8= 32

Note: In our cal we have cal 750 as well bcaz of first is 7 second digit is 5 and third is 0 - all three are different .so it is one possibility cal above.

Hence remember to subtract that from 32 -1=31 different digit numbers between 751 and 799

Different digit numbers between 800 to 899 . Cal same way as above

2*9*8 ( 2 bcaz we have only 8 or 9 possible in first place i.e hundreds place)
=144

Total 3 different digits between 751 to 999= 144+31= 175

Total - total 3 different digit numbers= numbers with at least two digit same
249-175=74


Ans choice D

Thanks you Bunuel ( learnt to solve this type of questions easily from your soln )

Hope this helps.
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