The security gate at a storage facility requires a fivE

4 * 3 * 5!/2!

= 12 * (5 * 4 * 3* 2!)/2!

= 12 * 60 = 720

Answer D

Hi bunuel,


I have a small doubt , why not we are considering that the first and last digit can occur in 2 ways and the middle 3 digits can occur in 6 ways.

for example middle 3 digits can be arranged in 3! ways among themselves after selecting 5c1 4c1 3c1 .

My question may be stupid, please correct my doubt.

Thanks

Consider this: two digit code XX, each digit must be distinct and can be 1, 2 or 3.

First digit can take 3 values (1, 2, or 3) and the second can take 2 values, total 3*2=6 codes:
12
13
21
23
31
32.

Similarly, for the original question: the first digit can take 4 values and the last digit can take 3 values, total 4*3. The same way for the middle 3 digits: the second digit can take 5 values (7 minus two we already used for first and last), the third 4 and the fourth 3.

Total: (4*3)*5*4*3=720.

Hope it's clear.

1. There is one large group i.e., digits 1 through 7, from which 5 digits should be selected.
2. The first and the last digits have a constraint that they should be odd. There is also a general constraint that each digit should be different.
3. Considering the above constraints the first and the last digits can be selected in \(4P2\) ways. 4 because there are 4 odd digits out of which they can be selected and because there should be no repetition it is \(4P2\) ways for those 2 positions.
4. The second digit cannot have the same digit as the first and the last digit. Therefore it can be chosen out of 7-2=5 digits. Similarly the third digit cannot have the same digit as the first, second and last. So it can be chosen in 4 ways and the fourth digit similarly in 3 ways.
5. So the number of lock codes possible is \(4P2 * 5* 4*3 = 720\)

Therefore the answer is D.

We have four odd numbers in this set 1, 3, 5, 7

First digit is 4C1 = 4
Last digit is 3C1 = 3 (we have to exclude the odd digit selected in the previous step.

mid digits are 3 of the remaining 5 digits (2, 4 , 6, in addition to two not selected odd digits) for the 2nd, 3rd and 4th digits in the code

mid digits 5C3 = 60

4*60*3= 720

Answer is D

Odd digits between 1 and 7
1 3 5 7 ==> 4
first and last place being fixed, can be filled with Odd number is
4* 5 *4* 3* 3

Total 720 combinations

Ans: D

Hi Bunuel,
Small clarification is required on this question.

Lets say.. First and last digits should be odd numbers . so it can be done in 5 ways (First) & 4 Ways (Last). Now, Since 5 digits are remaining for 3 places, ( 5C3 X 3!) can be done.... Finally answer is 360 ( can you explain?)

Dear rajendra00

Please take it this way...




So we have the following digits -

So, We have 4 digits to be filled in 2 places that can be done as -
Now we are left with 5 digits , ( 2 Odd Digits and 3 Even Digits ) and we now need to fill up 3 places using 5 digits that can be done as -
Now, Final Picture is
So, Total Number of ways is 4*5*4*3*3 = 720 ways..

Hence answer will be (D) 720..

Hope this helps.

Security gate requires a five digit lock code.
Lock code can take digits from 1 to 7 inclusive i.e. 1,2,3,4,5,6,7... Odd nos. among these nos are 1,3,5,7,total 4 in numbers.

Lets make boxes below to check in how many digits can be placed at the 5 different places of the lock code.

| 4ways (only odd allowed) | 5ways | 4ways | 3ways | 3ways (only odd allowed) |

Let me explain this..
At the first place we can place any of the 4 digits.
At the last place we can place any of 3 digits. (as digits can't be repeated.)

Now there are 5 digits left.
So, at the 2nd place we can place any of 5 digits.
at the 3rd place we can place any of 4 digits.
at the 4th place we can place any of 3 digits.

So, total number of lock codes possible = 4 * 5 * 4 * 3 * 3 = 80 * 9 = 720.

Answer D.

Hi,

I always get confused as to when to use the arrangement formula, i.e. n!, and when to use the combination formula, i.e. nCp.

In the given question I understood the selection of the 1st and last place.

My doubt is for the middle 3 places. Now since we have to 'choose 3 digits out of 5' to fill the middle 3 places, why can't we do 5C3?

Use arrangement formula n! only when u need to arrange the things like.

I am taking a very simple example for this .
If u want to make an arrangement of numbers 3,4,5 = 3! = 6
So lets see it by making the arrangement (345),(354),(534),(543)(453),(435) = Total 6 .

nCp is used when u have to make selection . For e.g. u need to select the 2 numbers from the three given numbers 3,4,5, = 3C2 = 3 ways.

Coming to the given question.
U have already understood the the first and last numbers,

Now for the middle places, We have to choose 3 digit out of 5 to fill the places. So it can be done in 5 x4 x3 ways = 60 ways

It can also be done in the way you suggested. Lets do it that way. So according to you choose 3 digits out of 5 to fill the middle 3 places = 5C3 = 5!/3!/2!
But these 3 numbers which you have selected out of 5 can also be arranged among themselves in 3! ways as explained in the example of 3,4,5. So you need to multiply your section with 3!. So the total ways become 5!/3!/2! * 3! = 5!/2! = 5*4*3 = 60 ways.. So answer comes to same ... either your way or my way.. Its just you need to think completely. Don't forget to appreciate with kudos if you like my answer.

I hope it solves your issue. Practice more questions to understand. With practice you will find that you don't need any formula and you can just calculate either way...

Let x-x-x-x-x be the 5-digit code.
No. of possibilities for each digit = 4-5-4-3-3 = 4*5*4*3*3 = 720 possibilities. Ans - D.

Thanks for the detailed explanation. Knowing the reasoning helps to apply the formula better.

Bunuel : Is it always necessary to start from the restrictions? Here first and the last digit for example. Or it is better to start from the restriction to avoid unnecessary error?

codes can be
odd= 1,3,5,7
even= 2,4,6
4*5*4*3*3 ; 720 IMO D

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