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Problem Solving - counting problems

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by anshumishra » Fri Jan 07, 2011 11:10 pm
koby_gen wrote:Of the three-digit integers greater then 750, how many have at least two digits that are equal to each other ?

(A) 56
(B) 70
(C) 72
(D) 74
(E) 78
Method 1 :

Total 3 digit numbers greater than 750 = 249
Total number of numbers having unique digits = Total no. of unique digits between 800 to 999 + Total no. of unique digits between 750 and 799
= 2*9*8 + 1*4*8 = 144 + 32 = 176 (This contains 750 as well)
Hence the numbers which have at least two digits equal = 249 - (176 -1) = 74.

Method 2 :

The numbers which have all the 3 digits equal (777,888,999) = 3
Lets calculate the numbers which have two digits equal.

First for the range (800-999) :
XXY 2*1*9 = 18
XYX 2*9*1 = 18
XYY 2*9*1 = 18

Total for this range = 3*18 = 54

For the range (750-799)
XXY 1*1*9 = 9
XYX 1*4*1 = 4
XYY 1*4*1 = 4

Total for this range = 17
Hence the numbers which have at least two digits equal = 54 + 17 + 3 = 74.
Last edited by anshumishra on Fri Jan 07, 2011 11:49 pm, edited 5 times in total.
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by Ramit88 » Fri Jan 07, 2011 11:18 pm
my ans = 74
what is OA?

My approach..
total numbers from 750 to 999 = 249 numbers

numbers between 800 and 999 with different digits = 2 x 9 x 8 = 144
numbers between 750 and 799 with different digits = 1 x 4 x 8 = 32

numbers with least two digits that are equal to each other = total numbers - numbers with different digits
= 250 - (144+32)
= 74



PS..sorry in case u dont understand my explanation..this is first time m trying to explain on this community..thnks
Last edited by Ramit88 on Fri Jan 07, 2011 11:36 pm, edited 1 time in total.
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by Ramit88 » Fri Jan 07, 2011 11:23 pm
anshumishra

I am getting a different answer. Here was the approach :

Method 1 :
Total numbers between 750 and 999 = 249
Total number of numbers having unique digits = Total no. of unique digits between 800 to 999 + Total no. of unique digits between 750 and 799
= 2*9*8 + 1*5*8 = 144 + 40 = 184

Hence the numbers which have at least two digits equal = 249 - 184 = 65.




u cannot take 5... its 4(5,6,8,9)..cant count 7
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by anshumishra » Fri Jan 07, 2011 11:26 pm
Ramit88 wrote:anshumishra

I am getting a different answer. Here was the approach :

Method 1 :
Total numbers between 750 and 999 = 249
Total number of numbers having unique digits = Total no. of unique digits between 800 to 999 + Total no. of unique digits between 750 and 799
= 2*9*8 + 1*5*8 = 144 + 40 = 184

Hence the numbers which have at least two digits equal = 249 - 184 = 65.

u cannot take 5... its 4(5,6,8,9)..cant count 7
Thanks Ramit, I just realized that :)
About to edit my post
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by Anurag@Gurome » Fri Jan 07, 2011 11:33 pm
koby_gen wrote:Of the three-digit integers greater then 750, how many have at least two digits that are equal to each other ?

(A) 56
(B) 70
(C) 72
(D) 74
(E) 78
Number of 3-digit integers with at least two same digits = (Number of 3-digit integers with exactly two same digits) + (Number of 3-digit integers with all the digits same) = (Number of 3-digit integers) - (Number of 3-digit integers with all three different digits)

Number of 3-digit integers greater than 750 = (999 - 750) = 249

Now let us calculate the number of 3-digit integers greater than 750 with all three digits different. The following cases are possible:
  • 1. 1st digit 7 --> 2nd digit may be 5, 6, 8, or 9 (4 possibilities) --> 3rd digit may be any one of the ten digits 0-9 except the two used earlier (8 possibilities) --> Total 4*8 = 32 integers. Now 750 is included in this calculation. Discarding that, total number of integers = 31

    2. 1st digit 8 or 9 (2 possibilities) --> 2nd digit may be any one of the ten digits 0-9 except the one used earlier (9 possibilities) --> 3rd digit may be any one of the ten digits 0-9 except the two used earlier (8 possibilities) --> Total 2*9*8 = 144 integers.
Therefore, number of 3-digit integers greater than 750 with all three digits different = (31 + 144) = 175

Number of 3-digit integers greater than 750 with at least two equal digits = (249 - 175) = 74

The correct answer is D.
Last edited by Anurag@Gurome on Fri Jan 07, 2011 11:46 pm, edited 1 time in total.
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by anshumishra » Fri Jan 07, 2011 11:42 pm
Anurag@Gurome wrote:
koby_gen wrote:Of the three-digit integers greater then 750, how many have at least two digits that are equal to each other ?

(A) 56
(B) 70
(C) 72
(D) 74
(E) 78
Number of 3-digit integers with at least two same digits = (Number of 3-digit integers with exactly two same digits) + (Number of 3-digit integers with all the digits same) = (Number of 3-digit integers) - (Number of 3-digit integers with all three different digits)

Number of 3-digit integers greater than 750 = (999 - 750) = 249

Now let us calculate the number of 3-digit integers greater than 750 with all three digits different. The following cases are possible:
  • 1. 1st digit 7 --> 2nd digit may be 5, 6, 8, or 9 (4 possibilities) --> 3rd digit may be any one of the ten digits 0-9 except the two used earlier (8 possibilities) --> Total 4*8 = 32 integers. Now 750 is included in this calculation. Discarding that, total number of integers = 31

    2. 1st digit 8 or 9 (2 possibilities) --> 2nd digit may be any one of the ten digits 0-9 except the one used earlier (9 possibilities) --> 3rd digit may be any one of the ten digits 0-9 except the two used earlier (8 possibilities) --> Total 2*9*8 = 144 integers.
Therefore, number of 3-digit integers greater than 750 with all three digits different = (31 + 144) = 175

Number of 3-digit integers greater than 750 with at least two equal digits = (249 - 175) = 72

The correct answer is C.
You meant D - 74 , right ?
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by Ramit88 » Fri Jan 07, 2011 11:43 pm
@ anurag sir

u said...Number of 3-digit integers greater than 750 with at least two equal digits = (249 - 175) = 72

but 249-175 =74
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by Ramit88 » Fri Jan 07, 2011 11:46 pm
why we are discarding 750..plz explain
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by Anurag@Gurome » Fri Jan 07, 2011 11:47 pm
Thanks guys.
Edited the reply.
Ramit88 wrote:why we are discarding 750..plz explain
Because 750 is NOT greater than 750.
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by anshumishra » Fri Jan 07, 2011 11:48 pm
Ramit88 wrote:why we are discarding 750..plz explain
Of the three-digit integers greater then 750, how many have at least two digits that are equal to each other ?
Thanks
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