For a certain alarm system, each code is comprised of 5 digi

hi abhishekj2512,

I have already tried this way and although it seems right, it doesn't give the right answer.
See below:

a. All 5 Distinct = 10C5 x 5! = 30240

b. 3 distinct, 2 of one kind = 10C3 X 7C1 X (5!/2!) = 8400

c. 2 of type1, 2 of type2, and a remaining solo = 10800

10800+8400+30240 = 49400

Another way that I considered was to calculate the complementary set:
10^5 - all possible codes, and deduct the number of ways to comprise a code with a digit that repeats 3 times, 4 times and 5 times. (It didn't work as well).

Seems like some error at your end, this comes out as 50,400
Using this, answer does come out correctly.

In this case, the forward and backward approach would involve same number of calculations.

you are right. I miscalculated the multiple factorial.

Thanks.

Sorry but I don't understand your answer. According to the question, no digit can be used more than twice. Then shall the answer be 10C5 = 30,240???
Please explain. Thanks

can someone help me here?
I do not understand why we are multiplying 5! here?
Select 5 digits out of 10 in 10C5 ways and arrange them in 5! ways which gives us 10C5x5! = 30240

i thought selecting 5 digits out of 10 would be just 10C5 .Am i missing anything?

Yes, the number of way to select 5 digits out of 10 is 10C5. But these 5 digits can be arranged in 5!=120 ways, each of which gives different code. Hence the number of codes for this case is 10C5*5!.

One could also do 10P5 (which is basically the same as 10C5*5!): selecting 5 digits out of 10, when the order of the selection matters.

Hope it's clear.

oh yess:).
Thanks Bunuel.

Responding to a pm:


As you can see, your case 3 has a problem.
There are 10 ways of writing the leftmost digit - correct
Then there are 10 ways of writing the next digit - correct
For the next digit, there could be 9 ways or there could be 10 ways depending on what was chosen previously:
e.g 22 __ __ __ - here there are 9 ways of choosing the next digit.
23 __ __ __ - here there are 10 ways of choosing the next digit. Both 2 and 3 can be repeated as opposed to case 2 where only one digit can be repeated.
Hence, don't do the question this way. Choose the repeated digits and then arrange as I have done above. Note that there will be far fewer cases here because 2 digits will be repeated so fewer different digits will be there. So the number of arrangements will be fewer.

Now think, why does case 2 work but case 3 does not.

Hi, could you please specify the logic for 2nd and 3rd cases.
2nd case:
I do not grasp the logic why you're counting 10C1 only once as long as we have assume that 2 digits are the same? And what is the rationale for using 5!/2! to arrange, why do we have a 2! in denominator? I'd rather used 3! to arrange remaining 3 distinct digits.
3d case:
Actually the same questions as above. I'd rather used 10C2 for the first pair of the same digits, then 8C2 for second pair of the same digits, and then 6C1 for the last one then arrange in 5!.
Would appreciate for explanation.
P.S. I'm not stable at computations. Some problems are solved smoothly however others like this one makes me stuck for many minutes...

You select one digit which has to be repeated. That digit will be written twice. So you just need to select one such digit. You can do that in 10C1 ways. The one selection could be 0 or 1 or 2 ...etc. This you will write twice.

Now you have place for 3 other digits. These you select in 9C3 ways.

How do you arrange something like 11234? Note that 12314, 21341 etc are all valid arrangements.
How do you arrange n things in which r are the same? You use n!/r!.

To know why, check out this video: https://www.youtube.com/watch?v=1zhltihi5VU

VeritasKarishma

Hi! I understand your solution however I solved the problem focusing on the available slots instead of available numbers. I managed to get the same number as you for the first two cases... however, case three I ended up with a different number although the equation we have is equivalent. Here's what I have:

Case one: ABCDE
number of different combinations 5 numbers can make out of 10 numbers
10P5=30240
Case two: AABCD
(# of different combinations 4 numbers can make out of 10 numbers) * (#of positions that one number of each code(A) that is selected to be repeated can take within the 5 slots)
10P4 * 5C2 = 50400

Case three: AABBC
(number of different combinations 3 numbers can make out of 10 numbers) * (#of positions that one number of each code(A) that is selected to be repeated can take within the 5 slots) * (#of positions that another number(B) of each code that is selected to be repeated can take within the remaining three slots)
10P3 * 5C2 * 3C2 = (10*9*8)*[5!/(3!2!)]*(3!/2!)=21600
=> what I have broken down is essentially the same equation as you have for case three, however I got 21600 for the answer.

Please shed some light on this!
Thank you!

In case 3, you have not selected the 2 digits that are to be repeated. Also use of 10P3 is incorrect since you are selecting 3 digits, not arranging them if you are selecting slots later.
Look, in such tricky questions, ALWAYS separate out the selection and arrangement.
Select 2 digits that are to be repeated first in 10C2 ways . Then select one more digit that needs to go with them in 8C2 ways. Then arrange all 5 in 5!/2!*2! ways.

why not the third case is 10C1 *9C1*8C1*5!/2!*2!

SOLUTION 1: TOTAL - NOTCASE
\(total(anydigit):10•10•10•10•10=10^5=100,000\)
\(22333(double,triple):10C1•9C1•arrangements(5!/2!3!)=10•9•10=900\)
\(33300(triple,distincts):10C1•9C2•arrangements(5!/3!)=10•36•20=7,200\)
\(44440(quad,distinct):10C1•9C1•arrangements(5!/4!)=10•9•5=450\)
\(55555(quintuple):10C1•arrangements(5!/5!)=10•1=10\)
\(TOTAL-NOT=100,000-(900+7200+450+10)=91,440\)

Answer (C)

SOLUTION 2: CASES
\(00000(distincts):10C5•arrangements(5!)=30,240\)
\(11000(one-pair,distincts):10C1•9C3•arrangements(5!/2!)=10•36•20=50,400\)
\(11220(two-pairs,distinct):10C2•8C1•arrangements(5!/2!2!)=45•8•30=10,800\)
\(CASES=30,240+50,400+10,800=91,440\)

Answer (C)

VeritasKarishma

Could you please help me with my query.

I see in your solution :-



Case 2 and 3, you have used 10c1 * 9c3 instead of 10c4 and 10C2 * 8C1 instead of 10C3.
Could you please explain why 10c4 and 10c3 will yield wrong answer? What was the logic behind 10c1 * 9c3 OR 10C2 * 8C1

Possible arrangements:

1. 5 Unique
Ex - 1 2 3 4 5

10C5 * 5! = 10*9*8*7*6
30240

2. 1 pair repeated & 3 unique
Ex - 1 1 0 2 3

10C1 * 9C3 * 5!/2!
10 * 9*8*7/(3*2) * 5*4*3
10 * 9*8*7 * 5*2
50400

3. 2 pairs repeated & 1 unique
Ex - 1 1 2 2 3

10C2 * 8C1 * 5!/(2!*2!)

5*9 * 8 * 5*3*2
10800


Sum it up:

30240+50400+10800
91440

Answer: (C)

rsrighosh

Look at case 2:

In our selection, of the 4 digits needed to make the five digit number, one digit is unique. That digit will be repeated. So to make 45722, we have to select a digit that will be repeated (which is 2 in our example) and 3 other digits (4, 5 and 7).

When we use nCr, out of n distinct elements, we select r. These r elements are all equivalent. But here, the 4 elements are not equivalent. One of them is special in that it will be repeated.
It is like selecting 1 manager and 3 workers. Out of 10 people, how will you select 1 manager and 3 workers? Will 10C4 work? No. You need to select the manager separately. The 3 workers can be selected as a group. So 10C1 * 9C3 will be done.

It is the same case here. We select the digit to be repeated separately.
When we do 10C4, how do we know which digit is repeated? Say we select 2, 4, 5, and 7. But the number could be 22457 or 24457 or 24557 or 24577 and their arrangements, each done in 5!/2! ways.

So this will be 10C4 * 4 = 10*9*8*7/4*3*2 * 4 = 10 * (9*8*7/3*2) which is the same as 10C1 * 9C3.