Question

· How many numbers formed with the digits
0, 1, 2, 5, 7, 8 will fall between 13 and 1000
if digits can be repeated?​

Answer 1

Answer:

207

Step-by-step explanation:

1) 2-digit  Number starting with  digit 1 and greater than 13.

15,17,18 :3

Remaining 2-digit numbers. (2,5,7,8)  (0,1,2,5,7,8).

_ _ : 4*6=24

2) 3-digit Numbers: (1,2,5,7,8)  (0,1,2,5,7,8) (0,1,2,5,7,8)

_ _ _ : 5*6*6=180

Total Numbers : 3+24+180 = 207.

Answer 2

Answer: 207

Step-by-step explanation:

We have digits 0 , 1 , 2 , 5 , 7, 8

Now for 2 digits greater than 13 starting with one are : 15, 17 ,18 = 3 ways  

Let the two digit number starting with 2 and more is _ _

Ones place can be filled by ( 0,1,2,5, 7, 8) 6 choices

Tens place can be filled by ( 2,5,7,8) 4 choices

Therefore total number of ways for two digits number = 6 x 4 = 24

Now the 3 digits number be _ _ _

Ones place can be filled by (0, 1, 2 ,5 ,7 ,8) 6 choices

Tens place can be filled by (0, 1, 2 ,5 ,7 ,8) 6 choices

Hundred place can be filled by (1, 2 ,5 ,7 ,8) 5 choices

Total number of ways  for 3 digit number = 5 x 6 x 6 = 180

Total = 3+ 24+ 180=  207

Hence, we can form 207 numbers using 0, 1, 2, 5, 7, 8 fall between 13 and 1000 with digits repeated .