QMS 202 Lecture 14: SXLBWnFR39Bi

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If you find my answer useful please put thumbs up. Here k= number of hats => 5 and n = number of balls => 30. S(k,n) = 1/k! j=1 5 ( "1)k "j (k j )( j )n: for exchanging the number they need two students so 15c2 = 15x14/ 2x1 = 105 ways they can answer the attendance for other students. 3 ( : 1 x1, x 444* g x 19, 1 10 + 8 + 4 x x s 2 2 5: . X s k s(n, k) = (-1)* (*))" k! 1 0: 1 1 30 x 244 x 28c, x27c 14 " 30 balls 5 hats case 1. 30c, x 24ch 28 10 9 18 q 84 x 44 8 7 2 . 29% 21. 4 x 19, . Case 4 x 5 5 6 30 24cc {\ce 12. 0 12 , 10 k s(n, k) = (-1)* (*))" k!

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