Permutations

S

sami123

New member
Joined
May 8, 2019
Messages
8
Four identical tins of peaches and six identical tins of pears are arranged in a row.calculate the number of different arrangements if the tin at each end contains the same type of fruits
 
I would add the number of arrangements with peaches at each end, and the number of arrangements with pears at each end.

Please show work so we can see where you need help, beyond the start.
 
Assume that there is a tin of peaches at each end. That leaves two tins of peaches and six tins of pears between. If all eight remaining tins were distinct there would be 8! ways to arrange them. But 2!= 2 of those are just rearrangements of the peaches and 6! of them are just rearrangements of the pears so there are \(\displaystyle \frac{8!}{2!6!}= 28\) ways to do this.

Assume that there is a tin of pears at each end. That leaves four tins of peaches and four tins of pears between. Do the same as above.
 
HallsofIvy said:
Assume that there is a tin of peaches at each end. That leaves two tins of peaches and six tins of pears between. If all eight remaining tins were distinct there would be 8! ways to arrange them. But 2!= 2 of those are just rearrangements of the peaches and 6! of them are just rearrangements of the pears so there are \(\displaystyle \frac{8!}{2!6!}= 28\) ways to do this.

Assume that there is a tin of pears at each end. That leaves four tins of peaches and four tins of pears between. Do the same as above.
Click to expand...
Thanks For Help
I also Posted One more question
Kindly Help In Solving this also
 
You must log in or register to reply here.
Top