11. Of the three-digit integers greater than 700, how many
have two digits that are equal to each other and the
remaining digit different from the other two?
(A)90
(B)82
(C)80
(D)45
(E)36
please explain in the simplest method
Answers
Answer:
(C)80
Step-by-step explanation:
Condition for 3-digit numbers: >700 and <999
Numbers will be considered in the format of:
7xx, 8xx, 9xx, 7x7, 8x8, 9x9, 77x, 88x, 99x where x is different digit
Each format counts 9 numbers except 700 which is excluded as per question.
So total count will be: 9*9-1=80
See below all numbers in the list to confirm the above.
701- 799
711, 722, 733, 744, 755, 766, 788, 799 = 8
707, 717, 727, 737, 747, 757, 767, 787, 797 = 9
770, 771, 772, 773, 774, 775, 776, 778, 779 = 9
800-899
800, 811, 822, 833, 844, 855, 866, 877, 899 = 9
808, 818, 828, 838, 848, 858, 868, 878, 898 = 9
880, 881, 882, 883, 884, 885, 886, 887, 889 = 9
900-999
900, 911, 922, 933, 944, 955, 966, 977, 988 = 9
909, 919, 929, 939, 949, 959, 969, 979, 989 = 9
990, 991, 992, 993, 994, 995, 996, 997, 998 = 9
So total = 8+8*9=80
Answer: Correct option is (C)80
Hope it helps
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