In how many ways can 10 identical presents be distributed among 6 children so that each child gets at least one present?
Answers
Answered by
1
no of identical presents = 10
no of children's to be distributed = 6
no of ways by which presents can be distributed = nPr = n!/(n-r)! × r!
= 10!/ (10-6)!×6!
=(10×9×8×7×6!)/4! × 6!
= (10×9×8×7)/(4×3×2)
= 10×3×7
=210
no of children's to be distributed = 6
no of ways by which presents can be distributed = nPr = n!/(n-r)! × r!
= 10!/ (10-6)!×6!
=(10×9×8×7×6!)/4! × 6!
= (10×9×8×7)/(4×3×2)
= 10×3×7
=210
Similar questions