Answer:
The correct option is B.
Explanation:
Total possible digits are 0,1,2,3,4,5,6,7,8,9.
Total number of digits in a credit card = 16
It is given that that the last digit of credit card is fixed.
Total number of possible ways for 16th digit = 1
U-Kan-Trust-Us Credit Card numbers begin with 61, 62, or 63.
Total number of possible ways for first two digits = 3
Remaining places = 16 - 1 - 2 = 13
In these 13 remaining places any of 10 digits can be occur.
Total number of possible ways for remaining 13 digits =
![10^(13)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/pc65kd4ihlidxin8dr2vu70y0p08ib5x9x.png)
The maximum number of credit cards that UKTU can issue is
![Total =3* (10^(13))* 1](https://img.qammunity.org/2020/formulas/mathematics/high-school/6vh5peisah8euvj4hu08gnhfv7zkvaiuzc.png)
![Total =3(10^(13))](https://img.qammunity.org/2020/formulas/mathematics/high-school/4n1p9w39zbv5ye2keax8mrs22enl9bhjjc.png)
Therefore the correct option is B.