We consider seven distinct boxes such that each box can contain at most seven items.

Romyn

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We consider seven distinct boxes such that each box can contain at most seven items.Seven distincts items will be distributed in these boxes.
A. In this part we assume that there is no order inside a box.
i. Find the probability that the seven items are placed in the same box.
ii. Find the probability that the seven items are placed in seven boxes.
iii. Find the probability that the seven items are placed in six boxes.
iv. Find the probability that two specifc items are placed in the same box.
B. In this part we assume that there is an order inside each box.
i. Find the probability that the seven items are placed in the same box.
ii. Find the probability that the seven items are placed in seven boxes.
iii. Find the probability that the seven items are placed in six boxes.
iv. Find the probability that two specifc items are placed in the same box.
 
Do you understand that the probability any given item is placed in a given box is 1/7?
Calling the boxes "A", "B", "C", "D", "E", "F", and "G", The probability all 7 items are placed in "A" is \(\displaystyle \frac{1}{7^7}\). Similarly, the probability all 7 items are placed in "B" is \(\displaystyle \frac{1}{7^7}\), etc. The probability all 7 items are placed in any one box is \(\displaystyle 7\left(\frac{1}{7^7}\right)= \frac{1}{7^6}\).
On the other hand, in order that the 7 items be placed in all 7 boxes, the probability an item is placed in "A" in 1/7. There are then 6 boxes left. The probability the next item is placed in a different box is 1/6. Then there are 5 boxes so the probability the third item is placed in yet a different box is 1/5. Continuing like that, the probability the 7 items are all placed in different boxes is \(\displaystyle \frac{1}{7}\frac{1}{6}\cdot\cdot\cdot \frac{1}{3}\frac{1}{2}\frac{1}{1}= \frac{1}{7!}\).

Now, at least try the other problems yourself and tell us what you have tried.
 
I have solved them all but only for when order matters
for part B i got :
1- 7/7^7
2- 7!/7^7
3- (6!x6)/7^7
4- (7x1x5^7)/7^7

but i have no idea how i should solve when order doesnt matter, can i just devide by 7!???

thank youuu HallsofIvy
 
Last edited:
Do you understand that the probability any given item is placed in a given box is 1/7?
Calling the boxes "A", "B", "C", "D", "E", "F", and "G", The probability all 7 items are placed in "A" is \(\displaystyle \frac{1}{7^7}\). Similarly, the probability all 7 items are placed in "B" is \(\displaystyle \frac{1}{7^7}\), etc. The probability all 7 items are placed in any one box is \(\displaystyle 7\left(\frac{1}{7^7}\right)= \frac{1}{7^6}\).
On the other hand, in order that the 7 items be placed in all 7 boxes, the probability an item is placed in "A" in 1/7. There are then 6 boxes left. The probability the next item is placed in a different box is 1/6. Then there are 5 boxes so the probability the third item is placed in yet a different box is 1/5. Continuing like that, the probability the 7 items are all placed in different boxes is \(\displaystyle \frac{1}{7}\frac{1}{6}\cdot\cdot\cdot \frac{1}{3}\frac{1}{2}\frac{1}{1}= \frac{1}{7!}\).

Now, at least try the other problems yourself and tell us what you have tried.
rn
 
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