Codey code !!!!!!

The problem says repetition does matter so that means you have to use the permutations formula...

$\frac{n!}{(n-r)!}$

And to help you decide which of the numbers is n and r you can use the term " Out of n terms pick r terms".So in his problem the $n$ is 7( or the number of numbers) and the $r$ is 4 because you have to choose 4 numbers to make the 4-digit code. And from there you can solve the equation...

$\frac{7!}{(7-4)!}$ ................ $\frac{5040}{6}$ ................. $5040/6=$ $\boxed{840}$

hm... let we see it..

• in the first digit we have 7 options 1,2,3,4,5,6,7
• 7 _ _ _
• because it cannot reused, we just have 6 options (if you choose 1 as the first digit you have 2,3,4,5,6,7)
• 7 6 _ _
• same like above you just have 5 options ( if you choose 2 as the second digit you have 3,4,5,6,7)
• 7 6 5 _
• 7 6 5 4.. so
• 7 * 6 * 5 * 4 = 840

There are 7 numbers to choose from in forming the four digit code.

For the first digit, you would have seven choices (any number from 1 to 7).

For the second digit, you would have 7 - 1 = 6 choices, since there is no repetition allowed. The number disclosed from this selection is the number chosen for the first digit, regardless of what the first digit was.

Similarly, you would have 7 - 2 = 5 choices for the third digit, excluding the first two distinct digits already chosen.

For the last digit, you would have 7 - 3 = 4 choices.

Take note that digit order would not matter (whether you choose the first or third digit of the code first) since each time you are already excluding the choice previously made.

As such, there would be 7 6 5*4 = 840 solutions possible.

The rationale for multiplying? Well, note that I have 7 choices for the first digit and 6 choices for the second.

For any of the 7 digits I picked for the first digit, I would still have 6 choices for the second digit. E.g. If I picked 4 for the first digit, I could choose 1,2,3,5,6,7 for the second digit. If the picked 7, I could choose 1,2,3,4,5,6 for the second digit.

I have seven possible ways to end up with 6 choices, thus the need for multiplication.

Doing the same for the third and fourth digits, the solution arrived at is 840 possible codes.

There are seven digits, but I only need four. So, I divided the factorial of the seven digits by the factorial of the seven digits minus the four gaps. It is as simple as: 7! / (7-4)! = 7! / 3! = 5040 / 6 = 840

n!/(n-r)! is the formula & n=7,r=4 ie., 7!/(7-4)!=5040/6=840

let we take permutations is easy 7p4 =840

to select a no. at unit place there are 7 ways and to fill next digit there are 6 ways similarly for next there are 5 ways and for last place there are 4 ways so total ways 7 6 5 4 = 840

There are 7 digits that can be used to form the 4-digit codes. To find the no. of such codes that can be formed, we need to find permutations of the 7 digits taken 4 at a time.

So, total no. of ways $=7P4 = 7\times 6\times 5\times 4 = 42\times 20 = \boxed{840}$

7 x 6 x 5 x 4 = 20 * 42 = 840. "n.(n-1).(n-2).(n-3)" or using Parmutation ---> 7P4 = 7!/(7-4)! = 7.6.5.4.3!/3! = 7.6.5.4 = 840. Answer : 840