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

Question

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?

A. 120
B. 240
C. 360
D. 480
E. 720

Ok, in the MGMAT guide on combinatronics they have two similar problems with & IMO they use two routes in solving them. This is frustrating me. I'm counting on the resident experts here at GMATclub to supply me with the defacto solution.

Bunuel · Accepted Answer

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.

cipher · Answer

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

TheSituation · Answer

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!!

Seryozha · Answer

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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aviddd · Answer

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

MathRevolution · Answer

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

Fdambro294 · Answer

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

CrackverbalGMAT · Answer

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.

Khwarizmi · Answer

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

Bunuel · Answer

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

Oppenheimer1945 · Answer

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

Lodz697 · Answer

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

Regor60 · Answer

Posted from my mobile device

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

KarishmaB · Answer

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

samarpan.g28 · Answer

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

7Amulya · Answer

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

rak08 · Answer

Bunuel  
whats wrong when i do this?

4c1 * 3c1 * 2! = odd ones
5c3*3! = others 
4*3*2*10*6

Krunaal · Answer

You can select the two odd terms together i.e. 4C2, and then arrange them in 2! ways.

When you select them one by one, like 4C1 and 3C1 and then multiply them you are already counting for arrangements, so 2! becomes redundant in that case.

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

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

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

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