AP EAPCET 08-Jul-2022 Shift 2 Paper

Total number of ways =[ Number of ways in which ₹10 can be distributed among three people]
- [Number of ways A can have more than ₹ 6]
[Number of ways B can have more than ₹ 7]
[Number of ways C can have more than ₹ 8]
Total possibilities =‌(n+r−1)Cn=‌(10+3−1)C10=‌12C10=‌

12!

10!2!

=66
A can have more than 6 in ‌5C3=10 ways
B can have more than 7 in ‌4C2=6 ways
C can have more than 8 in ‌3C1=3 ways
∴ Total number of ways =66−10−6−3=47