Suppose that a code consists of 5 digits, none of which is repeated. ( a digit is one of the 10 numbers 0,1,2,3,4,5,6,7,8,9.) how many codes are possible?

30240

Step-by-step explanation:

There are two primary ways to approach this problem - the logical and the mathematical.

The logical way is very simple:

Since you know that there are 5 digits and each must be different, start with the first digit and work your way up.

The first digit can be any of 10 digits (0–9).

However, the second digit can be any of only 9 digits as 1 digit has already been used and cannot be repeated.

Similarly, applying the same logic, there are only 8 options for the third digit, 7 for the fourth digit, and 6 for the fifth and final digit.

Now, multiply all these numbers together to get the total number of configurations like so:

10⋅9⋅8⋅7⋅6=30240

The more mathematically inclined way requires the use of a special formula. Since in this particular scenario, the order of the numbers matter, we can use the Permutation Formula:

P(n,r)=n!(n−r)!

where n is the number of numbers in the set and r

is the subset.

Since there are 10 digits to choose from, we can assume that n=10

.

Similarly, since there are 5 numbers that need to be chosen out of the ten, we can assume that r=5

.

Now, plug these values into the formula and solve:

=10!(10−5)!

=10!5!

=10⋅9⋅8⋅7⋅6

=30240

solution:

→c= a b , that is product of a and b is possible only when when a is m × n matrix and b is n × p matrix.

that is, number of columns of a= number of rows of b.

so, size of matrix , c= a b is , if a = (m × n)matrix

b= (n × p)matrix

then a b = [ (m × n)matrix] ×[   (n × p)matrix ]

→→c= a b= [m × p ] matrix

→→size of product a b will be equal to = number of rows of a × number of columns of b

what are ur choices?