In December 2024
Answer to Question #145300 in Statistics and Probability for Shiva n
slots.
(i) How many ways can you insert the 9 coins into the jukebox slots, if the order in which
the coins are inserted does not matter?
(ii) How many ways can you insert the 9 coins into the jukebox slots, if the order in which
the coins are inserted into each jukebox does matter?
(iii) How many ways can you insert 6 of the coins into one of the jukebox slots, if the order
in which the coins are inserted matters?
(i) We can make an equivalent formulation of the problem: Put 9 numbers into an arbitrary order (9! ways to do that due to the multiplication rule) and then split them into 4 parts (some parts can be empty, but all 4 parts must contain 9 numbers). The number of possible splittings was calculated manually. It is equal to "((9\\cdot8+8\\cdot7+7\\cdot6+6\\cdot5+5\\cdot4+4\\cdot3+3\\cdot2+2)=220" . The total number of ways to split 9 coins between 4 slots including the order is: "220\\cdot9!"
(ii) For each coin there are 4 different jukebox slots. We use the multiplication principle (https://www3.nd.edu/~apilking/Math10120/Lectures/Solutions/Topic03.pdf) and get "n=4^9" different ways to put coins into slots.
(iii) We assume that there is a fixed jukebox slot, in which we are going to put 6 coins. Using the multiplication principle, we get "6!=720" different ways.
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Comments
Dear Shiva n, You are welcome. We are glad to be helpful. If you liked our service, please press a like-button beside the answer field. Thank you!
Thanks you so much to the entire team. I am so thankful for your help, guidance and the way how you explain every concept that's a great work. Heartly thank you so much .
Dear Shiva n, the final answer to part (i) is 220*9!
Dear Shiva n, You are welcome. We are glad to be helpful. If you liked our service, please press a like-button beside the answer field. Thank you!
if i use (r+n-1)!/(r!(n-1)!) this formula which is according to you , then i am getting only 220 this answer . please check if i am wrong
if (r+n-1)!/(r!(n-1)!) this is formula for sub question 1 according to you and r = 9 and n = 4 then the final answer should be 220 Such as (9+4-1)! / ( 9!(4-1)! = 12! / 9! * 3! = that means (12*11*10*9!) / (9! * 3*2*1) here 9! will cancel and after division the we will get 220 at the end. please re-correct me if i am wrong anywhere .
Thanks you so much experts
The formula 9!/5!=9*8*7*6 in your comment does not work for the problem in the question because if 9 coins are inserted into one jukebox slot, then there is no coin to insert into the rest of jukebox slots. It will be the formula 9!/5!=9*8*7*6 for the number of ways when 4 of 9 coins are selected, the order matters, repetitions are not allowed.
In part (iii) if one jukebox is selected and 6 coins are inserted, then there are 6! ways to insert them if the order matters, repetitions are not allowed.
In part ii) the order matters, repetitions are allowed, there are no additionall restrictions, each jukebox can hold any number of coins, then permutations with repetition is applied and the formula is 4^9. There are 4 ways of putting a coin in the jukebox. As each jukebox can hold any number of coins, there are also 4 choices for each of the remaining coins 2, 3, 4, 5, 6, 7, 9. That is why the answer to (ii) is 4^9.
Dear Shiva n. It is coins that can be inserted into jukebox slots, not vice versa. In (i) the order does not matter, repetitions are allowed, there are no additional restrictions, then combinations with repetition is applied and the formula is (r+n-1)!/(r!(n-1)!)=(9+4-1)!/(4-1)!=12!/3!=12*11*10/6*9!=220*9!
Dear experts, if i use this approach for 1] condition does not matter ; n^r here n = 9 and r = 4 so 9*9*9*9 = 6561 . 2] condition does matter ; n!/(n-r)! here n = 9 r = 4 3] condition for does matter ; n!/(n-r)! here n = 6 r = 4 I know i am wrong in most of the cases so could you please clarify the formula with calculation , cause i am confused in condition or ques 1,2,3
Dear SHIVA N. In part i) there are 9 distinguishable coins and we can put 4-1=3 bars between the 9 coins, hence the total number of permutations of 9 coins and 4-1=3 bars is (9+4-1)!/(4-1)!=12!/3!=12*11*10/6*9!=220*9!. In part ii) each of 9 coins has 4 choices of taking, hence the total number is 4^9 (it is so-called number of r-permutations of n items with repetition). In part iii) the number of permutations of n objects without repetition was applied.
Still i am confused in solution , could you explain from basics or along with formula which is useful for this question?
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