A security company can use number 1 -7 to create a 5 digit

720 is the answer.

They use a "slot technique":

Ie. Spaces _,_,_,_,_

Space one has 4 options (1,3,5,7) and the last space has 3 (the 3 non-chosen odds from space one)

So we're sitting at 4,_,_,_,3

So between the first and last number been chosen 2 out of the 7 possibilites have been used leaving us with

4*5*4*3*3=720 combos

Ahhhhh, my problem on the related question is that I assumed the numbers were 1-9 when it was actually 0-9.

FRIG!!!

Thanks for the help guys!!

It's a relatively easy permutation problem that includes two restrictions. The order here matters because logically the code will 76543 will be different from 76534, so we possibilities for choice are high too.

This is a good question. Can someone please add options and provide OA?

Numbers to be used: 1, 2, 3, 4, 5, 6, 7

Lock: 5 digits: _ _ _ _ _

Condition: No numbers can be repeated and the first and last digit must be odd.

Odd numbers: 1, 3, 5, 7.

\(1^{st}\) place has 4 options and last place [\((5)^{th}\)] will have 3 options.

=> \(2^{nd}\) place will have: 5 options

=> \(3^{rd}\) place will have: 4 options

=> \(4^{th}\) place will have: 3 options

Total ways: 4 * 5 * 4 * 3 * 3 = 720

Answer: 720

I Always like to break these "Decisions" in Stages to make it clearer and easier to understand:


Available Digits: 1 --- 2 --- 3 -- 4--- 5 -----6 --- 7

Stage 1) We need to pick 2 of the 4 Odd Digits to be placed in the 1st and Last Spots - and then Arrange those 2 Selected Digits

-1- No. of Ways we can Choose 2 Odd Digits out of 4
4 - Choose - 2 = 4! / (2!2!) = 6 Ways

AND

-2- For Each Possible Group of 2 Digits, we can Arrange these 2 Digits between the 1st and 5th Spot in --- 2! Ways = 2 Ways


-------(6) * (2) = 12 Ways to Arrange the 1st and 5th Odd Digits


(Stage 2) For Each of the 12 Arrangements above, we will have 5 Digits remaining from which to Choose. We need to Choose 3 ---- and then find How Many Ways we can Arrange those 3 Chosen

-1- "5 - choose - 3" = 5! / (3!2!) = 10 Ways

AND

-2- For Each Group of 3 Digits Chosen, in how many ways can we Arrange the 3 Digits?

3! Ways = 6 Ways


-----(10) * (6) = 60 Ways to Choose and Arrange the Other 3 Digits



Fundamental Counting Principle: Stage 1 and Stage 2 are Events that happen one after another to form 1 Whole Scenario of a 5 Digit Code

(12 Ways to Choose and Arrange the 1st and 5th Digits) * (60 Ways to Choose and Arrange the Other 3 Digits) =

12 * 60 = 720 Different Codes


Answer - 720

A security company can use number 1 -7 to create a 5 digit lock combo. No numbers can be repeated and the first and last digit must be odd. How many unique lock combos can be created?

Odd numbers in the list :1,3,5,7

Conditions given:No numbers can be repeated and the first and last digit must be odd.

First digit can be filled by 4 ways as there are 4 odd numbers in the list.

Last digit can by filled by 3 ways because its given that no numbers can be repeated.

Out of 7 digits , two of them are used and we are left with five digits.


Second place can be filled by 5 ways .
Third and fourth places can by 4 and 3 ways respectively.

Total no of arrangements = 4 * 3 * 5 * 4 * 3 = 720 ways.

Bunuel this one popped up in the quiz and once again, there was no answer choice...

There are 0.01% questions with no OA's. We are working on it. Thank you!

1,2,3,4,5,6,7
1st and last digit={1,3,5,7} =4C2 2!
Rest all digits={o,o, 2,4,6} =5C3 3!

Total combos=12*10*6

In this case, I made it like this:

\(4C2 * 5 * 4 * 3 = 360\)

Considering that the first and last digits must be fixed odd numbers (4 out of 7).

What's my error?

Bunuel

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The two odd digits at the front and back can themselves be arranged in 2 ways

We can use the fundamental counting principle here. We have 5 empty slots in the password. We have 7 digits, 4 odd and 3 even.

O __ __ __ O

To fill in the first blank with an odd digit, we have 4 options.
To fill in the last blank with an odd digit, we have 3 options. (1 odd digit cannot be used)
To fill in the second blank, we have 5 options (2 leftover odd digits and 3 even digits)
To fill in the third blank, we have 4 options.
To fill in the fourth blank, we have 3 options.

Total ways = 4 * 3 * 5 * 4 * 3 = 720

Answer (E)

This counting principle is discussed here: https://youtu.be/LFnLKx06EMU

I think my approach will be hectic if the number set is large but here is how I did it.

First place can be filled in 4 ways and 5th place also be filled in 4 ways. Taking each 4 cases-

Case 1: choosing 1 at first place-
5th place can be filled in 3 ways ( 3, 5, 7). Now out of 7 numbers 2 have have been used. To fill the 2nd place we have 5 ways ( since repetition is not allowed and we have 5 numbers left out of 7); for 3rd place, 4 ways; for 4th place 3 ways.
Therefore, number of ways= 1*5*4*3*3= 180 ways.

Similarly,
case2: choosing 3 at 1st place; number of ways= 180
Case3: choosing 5 at 1st place; number of ways= 180
case4: choosing 7 at 1st place; number of ways= 180

Therefore, total number of ways= 180*4=720