How many 3 - digit even number can you form such that if one of the digits is 5, the following digit must be 7?

Case I: When no 5 is used
Last place can be occupied in 5 ways, the first place can be occupied in 8 ways. and second place in 9 ways.
so required no. of ways = 5*8*9 = 360
Case II , when 5 is used
5 has to be the first digit and it has to be followed by 7. So last digit will have 5 choices.
So there are only 5 such cases
Adding case I and case II, we can figure o5ut the total no. of ways which is 360+5 =365

350

11

5
572
574
576
578
570

365

240