64. There are 3 Indians and 3 Chinese in a group of 6 people. How many subgroups of thise group can
we choose so that every subgroup has at least one Indian?
(a) 56 (b) 52
(c) 48 (d) 44
The correct answer is (a) 56
Total subgroups of 6 people number 26=64, including the empty set. Subgroups with no Indians (only Chinese) number 23=8. Thus, subgroups with at least one Indian total 64−8=56.
Step-by-Step Solution
Label 3 Indians as I and 3 Chinese as C. Subgroups exclude the empty set but include singletons through the full group.
Calculate by cases based on Indians (k=1 to 3), with Chinese (m=0 to 3):
- k=1 Indian: 31×(23−1)=3×7=21 (subtract empty Chinese subset).
- k=2 Indians: 32×(23−1)=3×7=21.
- k=3 Indians: 33×23=1×8=8.
Total: 21+21+8=56.
Option Analysis
- (a) 56: Matches calculation; total non-empty minus all-Chinese non-empty (64−1−7=56).
- (b) 52: Incorrect; possibly excludes full group or miscalculates Chinese subsets.
- (c) 48: Incorrect; maybe assumes no all-Indians (56−8=48) or limits sizes.
- (d) 44: Incorrect; common error subtracting extra cases like all-Indians twice.
Introduction to 3 Indians 3 Chinese Subgroups Problem
The “3 Indians 3 Chinese subgroups at least one Indian” question tests combinations in competitive exams like GATE. With 6 people (3I + 3C), find subgroups ensuring at least one Indian. Total power set is 26=64; subtract 8 all-Chinese for 56.
Detailed Calculation Method
Use complementary counting for efficiency:
- Total subsets: 26=64.
- No Indians: 23=8.
- Valid: 56.
Case breakdown confirms:
| Indians (k) | Ways to Choose Indians | Chinese Subsets | Subtotal |
|---|---|---|---|
| 1 | 31=3 | 23=8 | 24 |
| 2 | 32=3 | 23=8 | 24 |
| 3 | 33=1 | 23=8 | 8 |
Total: 56
Common Errors in Options
Options like 52 or 48 arise from excluding the empty set prematurely or all-Indians. Always verify with both methods.
GATE Exam Tips
Practice power set subtraction for “at least one” constraints. Similar to GATE ECE 2017.
