Since in a pass code, the placement of the digits is
important, therefore this means that to solve for the total number of
possibilities we have to make use of the principle of Permutation. The formula
for calculating the total number of possibilities using Permutation is given
as:
P = n! / (n – r)!
where,
n = is the total amount of numbers to choose from = 20
r = is the total number of digits needed in the passcode =
4
Therefore solving for the total possibilities P:
P = 20! / (20 – 4)!
P = 20! / 16!
P = 116,280
<span>Hence there are a total of 116,280 possibilities of pass
codes.</span>
Answer:
0.29
Step-by-step explanation:
7x=2
/7 /7
x=0.29
-hope it helps
I believe the answer to your question is b.
Let's set up some simultaneous equations. Let the number of nickels be n, and the number of quarters be q. Now I'm from England so I had to look up the value of a nickel (I could guess the quarter though to be 25c), but apparently it's 5c
Okay we know that the total value of the coins is 150. This gives:
25q + 5n = 150
Rearranging to make n the subject:
n = 30 - 5q
Next, we use the second statement:
n = 2q - 5
Substituting this into the first equation:
2q - 5 = 30 - 5q
7q = 35
q = 5
Putting this value for q into the first equation:
n = 30 - 5(5) = 5
Hence he has 5 of each coin. I hope this helps :)
