Question

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​

Answer 1

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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