98.1k views
1 vote
(a) What is the probability that an integer between 1 and 10,000 has exactly three 5's and one 3? (b) How many ways are there to distribute 50 identical jelly beans among six children if each child must get at least one jelly bean? (c) How many ways are there to distribute 21 different toys among six children (Alex, Ella, Jacqueline, Kelly, Rob, Stephen), if two children gets 6 toys, three children get 2 toys and one child get 3 toys? (d) How many "words" can be formed by rearranging INQUIRING (3 I's, 2 N's, 1 Q, 1 U, 1 R, 1G) so that U does not immediately follow Q? (e) If a person owns 6 mutual funds (each with at least one stock), where (i) these mutual funds together have a total of 61 stocks and (ii) the largest fund is Zillow, what is (A) the smallest number of stocks in Zillow and (B) the largest number of stocks in Zillow?

1 Answer

5 votes

Answer:

(a) To find the probability that an integer between 1 and 10000 has exactly three 5's and one 3, we need to count the number of such integers and divide by the total number of integers between 1 and 10000. There are 4 positions in the integer that need to be filled with 3 5's and 1 3, so we can count the number of ways to choose these positions (which is C(4,1) = 4) and the number of ways to fill them with the 5's and 3 (which is 2 * 2 * 2 = 8), and then count the number of ways to fill the remaining positions with digits other than 5 and 3 (which is 8 * 8 * 8 * 8 = 4096). Therefore, the total number of integers between 1 and 10000 with exactly three 5's and one 3 is 4 * 8 * 4096 = 131072, and the probability of selecting such an integer is 131072/10000 = 131/10,000.

(b) To distribute 50 identical jelly beans among six children so that each child gets at least one jelly bean , we can use the stars and bars method. We place 5 bars among the 50 jelly beans to divide them into 6 groups, and we choose the positions of the bars from the 49 spaces between the jelly beans (since the first and last spaces cannot be used). There are C(49,5) ways to do this, which is approximately 1.47 * 10^9.

(c) To distribute 21 different toys among six children according to the given conditions, we can consider the number of toys received by each child separately. Two children get 6 toys each, so we can choose the two children in C(6,2) ways and the toys for each child in C(21,6) ways, so the total number of ways to distribute 12 toys among two children is C(6,2) * C(21,6)^2. Similarly, three children get 2 toys each, so we can choose the three children in C(6,3) ways and the toys for each child in C(15,2) ways, so the total number of ways to distribute 6 toys among three children is C(6,3) * (C(15,2))^3. Finally, one

Step-by-step explanation:

User JohnSUN
by
7.5k points

No related questions found