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The number of groups of three or more distinct numbers that can be chosen from 1, 2, 3, 4, 5, 6, 7 and 8 so that the groups always include 3 and 5, while 7 and 8 are never included together is:
A) 45
B) 47
C) 48
D) 60
E) 63
We need to select 3 or more numbers from the numbers 1 to 8 such that 3 and 5 are definitely there.
The remaining numbers are: 1, 2, 4, 6, 7, 8.
We need to select 1 or more numbers from these.
We cannot select both 7 and 8.
Case 1: Let us select 7. We need to remove 8 and from the remaining 4 numbers, we can select none, one, two, three or all.
The number of ways = \(2^4\) = 16
Case 2: Let us select 8. We need to remove 7 and from the remaining 4 numbers, we can select none, one, two, three or all.
The number of ways = \(2^4\) = 16
Case 3: Let us remove 7 and 8. From the remaining 4 numbers, we can select one, two, three or all.
The number of ways = \(2^4 - 1\) = 15
Thus, total number of ways = 16 + 16 + 15 = 47
Answer BWatch an
Alternative approach in the YouTube video here: