We have 7 numbers: 1-2-3-4-5-6-7, out of which 4 are odd. Want to make 5 digit code, plus first and last digits must be odd: OXXXO.

#1:
\(P^2_4*P^3_5=12*60=720\)

\(P^2_4\) - choosing 2 odds out of 4 odds, (note that order matters in this case: 1---3 is different from 3---1);

\(P^3_5\) - choosing 3 numbers (for XXX) out of 5 left, (order matters: o234o is different from o342o).

OR #2:
4 choices for the first slot, 3 choices for the last slot (4 odds-1=3), 5 choices for second slot (7 numbers -2 odds), 4 choices for the third slot and 3 choices for the fourth slot: 4*3*5*4*3=720.

OR #3:
\(C^2_4*2!*C^3_5*3!=720\), this is basically the same as the first approach, but here we choose 2 odds out of 4 and then arranging them (2!) and respectively we choose 3 other numbers from 5 left and then arranging them (3!).

Hope it's clear.
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Lets take it this way, first and last digit needs to be odd so that means it take 1,3,5,7

So, for first digit we have 4 options and for last digit we have 3 digits

for the rest of the 3 digits we have 5 digits to choose from so \(5C3= 10\) options

Total combinations = 4 * 3 * 10 = 120

Cheers

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.

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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
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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