A father with eight children takes three at a time to the zoological garden,…
2025
A father with eight children takes three at a time to the zoological garden, as often as he can without taking the same three children together more than once. How often will he go and how often will each child go?
- A.
56, 35
- B.
92, 42
- C.
56, 21
- D.
56, 42
Show answer & explanation
Correct answer: C
Concept: The number of ways to choose r items from n distinct items, ignoring order, is given by the combination formula C(n, r) = n!/(r!(n-r)!). When each such selection is one trip, the total number of trips equals C(n, r); to find how often one particular item appears across all selections, fix that item and count the ways to fill the remaining (r-1) slots from the other (n-1) items -- i.e., C(n-1, r-1).
Application:
Total trips = number of distinct 3-child groups from 8 children = 8C3 = 8!/(3! x 5!) = (8 x 7 x 6)/(3 x 2 x 1) = 56.
Fix one particular child; the remaining 2 seats in that child's group are filled from the other 7 children = 7C2 = 7!/(2! x 5!) = (7 x 6)/(2 x 1) = 21.
So the father makes 56 trips in total, and each child accompanies him on 21 of those trips.
Cross-check: Each of the 56 trips carries 3 children, so total child-appearances = 56 x 3 = 168. With 8 children sharing these appearances equally by symmetry, per-child appearances = 168 / 8 = 21 -- matching the value found by fixing one child directly.
Result: The father goes 56 times in total, and each child accompanies him on 21 of those trips.