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
(A) 56
(B) 70
(C) 72
(D) 74
(E) 78
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Method 1 :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
Thanks Ramit, I just realized thatRamit88 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
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)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
You meant D - 74 , right ?Anurag@Gurome wrote: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)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 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:Therefore, number of 3-digit integers greater than 750 with all three digits different = (31 + 144) = 175
- 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.
Number of 3-digit integers greater than 750 with at least two equal digits = (249 - 175) = 72
The correct answer is C.
Because 750 is NOT greater than 750.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 ?Ramit88 wrote:why we are discarding 750..plz explain