By Marcel B. Finan

**Read or Download A Probability Course for the Actuaries: A Preparation for Exam P 1 PDF**

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**Extra resources for A Probability Course for the Actuaries: A Preparation for Exam P 1**

**Example text**

Proof. 1: There are C(n, n1 ) possible choices for the first box; for each choice of the first box there are C(n − n1 , n2 ) possible choices for the second box; for each possible choice of the first two boxes there are C(n − n1 − n2 ) possible choices for the third box and so on. 1, there are C(n, n1 )C(n−n1 , n2 )C(n−n1 −n2 , n3 ) · · · C(n−n1 −n2 −· · ·−nk−1 , nk ) = different ways n! n2 ! · · · nk ! 3 How many ways are there of assigning 10 police officers to 3 different tasks: • • • Patrol, of which there must be 5.

Then the number of solutions to the given equation is 11 + 3 − 1 11 = 78. 9 How many solutions are there to the equation n1 + n2 + n3 + n4 = 21 where n1 ≥ 2, n2 ≥ 3, n3 ≥ 4, and n4 ≥ 5? 52 COUNTING AND COMBINATORICS Solution. We can rewrite the given equation in the form (n1 − 2) + (n2 − 3) + (n3 − 4) + (n4 − 5) = 7 or m1 + m2 + m3 + m4 = 7 where the mi s are positive. The number of solutions is C(7+4−1, 7) = 120 Multinomial Theorem We end this section by extending the binomial theorem to the multinomial theorem when the original expression has more than two variables.

Solution. The three remaining people not chosen can be viewed as a third “committee”, so the number of choices is just the multinomial coefficient 10 4, 3, 3 = 10! 3! Distributing n Identical Objects into k Boxes Consider the following problem: How many ways can you put 4 identical balls into 2 boxes? We will draw a picture in order to understand the solution to this problem. Imagine that the 4 identical items are drawn as stars: {∗ ∗ ∗∗}. 50 COUNTING AND COMBINATORICS If we draw a vertical bar somewhere among these 4 stars, this can represent a unique assignment of the balls to the boxes.