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

Question

Brittany Bartlett

Mathematics

30 November, 16:22

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

+1

Answers (1)

Know the Answer?

Howard · Answer 1

80.

Step-by-step explanation:

Three digit number can only have three patterns

A. All digits are distinct

B. Two digits are same and third is different

C. All three digits are alike.

We need to calculate B

that is B = Total - A-C

Total numbers from 700 to 999 = 299.

Number numbers having all digits distinct = 3*9*8 = 216 (first digit can have only three digits 7,8,9)

All three are alike = 3 (777,888,999).

So, 299 - 216-3 = 80.

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) 90B) 82C) 80D) 45E) 36

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