Set up a system of equations.
0.10d + 0.25q = 39.25
d + q = 250
Where 'd' represents the number of dimes, and 'q' represents the number of quarters.
d + q = 250
Subtract 'q' to both sides:
d = -q + 250
Plug in '-q + 250' for 'd' in the 1st equation:
0.10(-q + 250) + 0.25q = 39.25
Distribute 0.10:
-0.10q + 25 + 0.25q = 39.25
Combine like terms:
0.15q + 25 = 39.25
Subtract 25 to both sides:
0.15q = 14.25
Divide 0.15 to both sides:
q = 95
Now plug this into any of the two equations to find 'd':
d + q = 250
d + 95 = 250
Subtract 95 to both sides:
d = 155
So there are 95 quarters and 155 dimes.
Step-by-step explanation:
Permutation
:The number of ways to choose a sample of r elements from a set of n distinct objects where order does matter and replacements are not allowed.

Factorial
: There are n! ways of arranging n distinct objects into an ordered sequence.
Considering a situation when n = r in a permutation, nPr reduces to n!, a simple factorial of n.
Proof: 3P3 = 3!
n = 3 and r = 3

But 0! = 1
3P3 = 3!
Answer: 24.34 is his balance.
Step-by-step explanation:
.95 in decimal form and 95/100 as a fraction or 19/20