How many three digit numbers greater than 700 exist if, two

Question

How many three digit numbers greater than 700 exist if, two of the digits in the number are the same and the third is different from the other two?

A. 45
B. 67
C. 81
D. 80
E. 48

OA/OE by the end of the day...

Yurik79 · Answer

broot force
711 
722
733
744
755
766
788
799
-----
8

770
771
772
773
774
775
776
778
779
----
9

800
811
822
833
844
855
866
877
899
-----
9
880
881
882
883
884
885
886
887
899
----
9
900
911
922
933
944
955
966
977
988
----
9
total 53 ((I don't have any idea where am I wrong

MA · Answer

total = 999- 700 = 299
diffrent three digits = 3 (9x8) = 216
same three digits = 3
so two same digits = 299 - 216 - 3 = 80

haas_mba07 · Answer

MA,
 Can you please explain? 
As I understand this problem, we can split it into 3 different sets:

a). All #s 7xx
 The hundreds digit is always 7 (as > 700)
There are two cases:
i) Tens digit is equal to 7 : only 1 way
 Units digit can be filled in 8 ways (excluding 7)

ii) Units digit is equal to 7 : only 1 way
 Tens digit can be filled in 8 ways (excluding 7)

Total number of ways = 1x1x8 + 1x1x8 (case i +case ii)
= 16
As > 700, exclude 700 in the count = 15

b). All #s 8xx
 Same logic as above; # of ways = 16

c). All #s 9xx
 Same logic as above; # of ways = 16

Total numbers = 16 + 16 + 15 = 32+15 =47

I know I missed something, but not sure what?

laxieqv · Answer

well, i'm not good at probability but lemme try this

No of 3-digit number > 700 = 3*10*10 - 1 ( the first digit can be formed by 7,8,9 . Minus 1 because it's when the number is 700) = 299 No of 3-digit number> 700 in which 3 digits are the same= 3 ( 777,888,999) No of 3-digit number> 700 in which 3 digits are different= 3*9*8= 216 ---> the number of 3-digit number which satisfies the condition = 299-3-216= 80

MA · Answer

alternatively:

1. for 7xx: 
701-710 = 707
711-720 = 711 and 717 
721-730 = 722 and 727
731-740 = 733 and 737
741-750 = 744 and 747
750-760 = 755 and 757
761-770 = 766, 767 and 770. 
771-779 = 8 except 777.
780-789 = 787 and 788
790-799 = 797 and 799.
total 26.

2. for 8xx: same as above except 1 more for 800. so total = 27.
3. for 9xx: same as above except 1 more for 900. so total = 27.

so total = 26+27+27 = 80.

now you can see above what you missed.

ps_dahiya · Answer

7XX = 10 - 1 (Exclude 777) = 9
77X = 10 - 1 (Exclude 777) = 9
7X7 = 10 - 1 (Exclude 777) = 9
same way for 800 and 900.

We have 27* 3 = 81
But subtract 700. So answer = 81-1 = 80

Futuristic · Answer

D. 

Case 1:

7xx exclude 0 and 7 = 8
77x exclude 7 = 9
7x7 exlclude 7 = 9

Similarly for 8, we have 8xx, 8x8, 88x, and we don't need to exclude 0 so its 9 for each case, giving a total of 27
Identical for 9, we get 27

total = 26+27x2 = 80

prude_sb · Answer

81 ? 

hundered's digit can be 7/8/9

case1: 7

select 7 for 1s or 10s digit and not have 7 for the other digit 2*9
select 2 identical numbers...9 numbers (cos 7 cant be included) so 9

so total 3*9

simiilarly for 8 and 9 so 3*3*9 = 81 ?

prude_sb · Answer

and > 700 so 700 cant be there ...so 80

haas_mba07 · Answer

OA is D. 80.

As all of you other than me got the answer.. I need to go back and understand my solution.

Thanks all/

defenestrate · Answer

Or you could look at the answers and guess that it is either 80 or 81 then look at the question, see the word greater and understand that the trick here is that, and it is 80 :shock:

Futuristic · Answer

This might work for you, but if you can solve the problem, why guess?

How many three digit numbers greater than 700 exist if, two

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