Respuesta :
We have 90 coins, so we have 90! ways in arranging them in 90 slots.
Since we have identical coins, we have overcounted by a factor of 20!, 30!, and 40!
We can visualise this with a portion of the coins: let's take 3 pennies, 4 nickels, and 5 dimes to demonstrate this.
We'll call them p (pennies), n (nickels), and d (dimes).
P₁ P₂ P₃ N₁ N₂ N₃ N₄ D₁ D₂ D₃ D₄ D₅
Without the superscripts, we get the arrangement: PPPNNNNDDDD
This is one arrangement.
Now, let's try to change the position of the coins.
P₂ P₁ P₃ N₂ N₁ N₄ N₃ D₂ D₁ D₃ D₅ D₄
This is another arrangement, since the order matters. So, without the superscripts, we get:
PPPNNNNDDDDD
But we just made this arrangement in the example before.
In fact, we have this many duplicates, 20!30!40! times because the coins can change the order without us noticing.
Thus, we need to exclude these repetitions from our final number:
[tex]P = \frac{90!}{20! \cdot 30! \cdot 40!}[/tex]
Since we have identical coins, we have overcounted by a factor of 20!, 30!, and 40!
We can visualise this with a portion of the coins: let's take 3 pennies, 4 nickels, and 5 dimes to demonstrate this.
We'll call them p (pennies), n (nickels), and d (dimes).
P₁ P₂ P₃ N₁ N₂ N₃ N₄ D₁ D₂ D₃ D₄ D₅
Without the superscripts, we get the arrangement: PPPNNNNDDDD
This is one arrangement.
Now, let's try to change the position of the coins.
P₂ P₁ P₃ N₂ N₁ N₄ N₃ D₂ D₁ D₃ D₅ D₄
This is another arrangement, since the order matters. So, without the superscripts, we get:
PPPNNNNDDDDD
But we just made this arrangement in the example before.
In fact, we have this many duplicates, 20!30!40! times because the coins can change the order without us noticing.
Thus, we need to exclude these repetitions from our final number:
[tex]P = \frac{90!}{20! \cdot 30! \cdot 40!}[/tex]
The number of different ways one can put all the coins in a row is [tex]\frac{90!}{20!30!40!}[/tex].
It is given that
Number of pennies = 20
Number of nickels = 30
Number of dimes = 40
What is the number of ways n items can be arranged in a row?
The number of ways n items can be arranged in a row is n!.
So, the total number of ways one can put all the coins in a row = (20+30+40)! = 90!
Since, pennies, nickels, and dimes are identical. This means there are some ways that are repeating or we can see they are over-counted.
So, we need to divide 90! by (20!30!40!) to get a number of different ways to put the coins in a row.
Therefore, The number of different ways one can put all the coins in a row is [tex]\frac{90!}{20!30!40!}[/tex].
To get more about permutation visit?
https://brainly.com/question/1216161
