Questions regarding permutations .

[sankalpmittal]

Questions regarding "permutations".

Homework Statement

These two mini questions , I think , can be adjusted in one thread itself.

1. How many total 5-digit numbers divisible by 6 can be formed using 0,1,2,3,4,5 if repetition of digits is not allowed ?

2. In the decimal system of numeration , find the number of 6-digit numbers in which the digit in any place is greater than the digit to the left of it.

Homework Equations

Concepts of permutations...

The Attempt at a Solution

Ok , so for the first one ,

I made the cases as follows...

For the divisibility by 6, the number should be divisible by 3 as well as 2.

Case 1: Fix "0" at the end. The the sum can be 1+2+3+4+5 = 15
I have to fill like this : __ __ __ __ 0
Here 4 places have to be filled such that the sum yields the multiple of 3. If I subtract a number say x from 15 , I must get a multiple of 3 which is possible when x=3.
Thus Permutations = 4! = 24

Case 2 : Fix "2" at the end.
__ __ __ __ 2
The sum can be : 1+0+3+4+5 = 13
I must have , 13-x = a multiple of 3
So , x=1 , or x= 4

Thus , permutations : 3x3x2x1 = 18
As there are two sub cases , so Required permutations :18x2 = 36

Case 3 : Fix "4".
__ __ __ __ 4
The sum can be : 1+2+3+0+5 = 11
I must have , 11-x = a multiple of 3
So , x=2 , x=5

Thus here also Permutations = 36 (as there are two sub cases)

Thus total Permutations : 36+36+24 = 96 , but answer given is 108 ! How ?

For second one , I attempted for over 3 pages but failed ! I don't know how to do this !

Please help !

Thanks in advance...

 

[HallsofIvy]

Homework Statement

These two mini questions , I think , can be adjusted in one thread itself.

1. How many total 5-digit numbers divisible by 6 can be formed using 0,1,2,3,4,5 if repetition of digits is not allowed ?

2. In the decimal system of numeration , find the number of 6-digit numbers in which the digit in any place is greater than the digit to the left of it.

Homework Equations

Concepts of permutations...

The Attempt at a Solution

Ok , so for the first one ,

I made the cases as follows...

For the divisibility by 6, the number should be divisible by 3 as well as 2.

Yes, and that means that the digits must sum to a multiple of 3 and the last digit must be even- 0, 2, or 4.

[/quote]Case 1: Fix "0" at the end. The the sum can be 1+2+3+4+5 = 15
I have to fill like this : __ __ __ __ 0
Here 4 places have to be filled such that the sum yields the multiple of 3. If I subtract a number say x from 15 , I must get a multiple of 3 which is possible when x=3.[/quote]
Okay, so the first four digits must be 1, 2, 4, 5, in any order.

Thus Permutations = 4! = 24

Right.

Case 2 : Fix "2" at the end.
__ __ __ __ 2
The sum can be : 1+0+3+4+5 = 13
I must have , 13-x = a multiple of 3
So , x=1 , or x= 4

So with x= 1, the four digits must be 0, 3, 4, 5 but 0 cannot be the leading digit so there are 3(3!)= 18 as you say. With x= 4, the four digits must be 0, 1, 3, 5 and the same as before totaling 36.

Thus , permutations : 3x3x2x1 = 18
As there are two sub cases , so Required permutations :18x2 = 36

Yes.

Case 3 : Fix "4".

as last digit.

__ __ __ __ 4
The sum can be : 1+2+3+0+5 = 11
I must have , 11-x = a multiple of 3
So , x=2 , x=5

Thus here also Permutations = 36 (as there are two sub cases)

Thus total Permutations : 36+36+24 = 96 , but answer given is 108 ! How ?

I don't believe you have done anything wrong. I think 96 is the correct answer.

For second one , I attempted for over 3 pages but failed ! I don't know how to do this !

Please help !

Thanks in advance...

Looks like pretty much "brute strength". If the first digit is 1 (this is the "hard" case), the second digit can be any of 0, 2, 3, 4, or 5. It cannot be any larger than 5 because then there won't be four larger digits to follow. If the second digit is 2, the third digit can be any of 3, 4, 5, or 6. Etc.

 

[sankalpmittal]

Looks like pretty much "brute strength". If the first digit is 1 (this is the "hard" case), the second digit can be any of 0, 2, 3, 4, or 5. It cannot be any larger than 5 because then there won't be four larger digits to follow. If the second digit is 2, the third digit can be any of 3, 4, 5, or 6. Etc.

Ok , but I have tried the so called "brute strength" , and have already filled 3-4 pages of my copy. It becomes too long.

Can you suggest me any other way ? We get only 2-3 minutes of time in exam to solve such questions.

 

[I like Serena]

Hi sankalp!

Thus total Permutations : 36+36+24 = 96 , but answer given is 108 ! How ?

I agree with HoI, I believe 96 is the correct answer.

For second one , I attempted for over 3 pages but failed ! I don't know how to do this !

Suppose you pick 6 different digits and sort them.
How many possibilities?
And how many of those start with a zero? Because those won't count.

 

[ehild]

You need to make five-digit numbers, so one of the six is not used. The number is divisible by three, so the sum of its digits must be divisible by 3. 0+1+2+3+4+5 = 15 is divisible, you can only left out 0 and 3.

If you choose 1,2,3,4,5, 2 or 4 can be on the last place. How many possibilities you have to fill the other four places?

If you choose 0,1,2,4,5, 0 the last digit should be 0, 2 or 4. Three cases to investigate.


ehild

 

[sankalpmittal]

Hi sankalp!
I agree with HoI, I believe 96 is the correct answer.

Hii ILS !

(It has been a long time right ?)

Alright may be there is a misprint in the answer given in my booklet for the first question , I asked here.

Suppose you pick 6 different digits and sort them.
How many possibilities?

¹⁰C₆ right ? Because we have sorted all the permutations.

And how many of those start with a zero? Because those won't count.

So if I leave 0 , the combinations become ⁹C₆.
And yes ⁹C₆ is the required answer !

Thanks ILS !

@ehild

You need to make five-digit numbers, so one of the six is not used. The number is divisible by three, so the sum of its digits must be divisible by 3. 0+1+2+3+4+5 = 15 is divisible, you can only left out 0 and 3.

If you choose 1,2,3,4,5, 2 or 4 can be on the last place. How many possibilities you have to fill the other four places?

If you choose 0,1,2,4,5, 0 the last digit should be 0, 2 or 4. Three cases to investigate.ehild

Hii ehild !
I already did the first question , and it has been point out correct by Halls of Ivy and ILS.

 

[haruspex]

Hii ehild !
I already did the first question , and it has been point out correct by Halls of Ivy and ILS.

Except, Halls of Ivy and ILS both agreed with your 96. Following ehild's analysis (with which I agree) 108 is right: 48 leaving out 0, plus 72 leaving out 3 and allowing a leading zero, minus 12 with leading zero.

 

[ehild]

Except, Halls of Ivy and ILS both agreed with your 96. Following ehild's analysis (with which I agree) 108 is right: 48 leaving out 0, plus 72 leaving out 3 and allowing a leading zero, minus 12 with leading zero.

Yes, 108 is the correct result. Leaving out zero, and putting 2 or 4 as the last digit, you get 4! possibilities for each, that is 2*24 = 48.

Leaving out 3 the last digit can be 0 and the permutations of 4 for the other digits. Or the last digit can be either 2 or 4. Then you can choose 3 numbers as the first digit (0 is excluded) and 3! possibilities for the other digits: 4!+2*3*3! = 60

ehild

 

[sankalpmittal]

Yes, 108 is the correct result. Leaving out zero, and putting 2 or 4 as the last digit, you get 4! possibilities for each, that is 2*24 = 48.

Leaving out 3 the last digit can be 0 and the permutations of 4 for the other digits. Or the last digit can be either 2 or 4. Then you can choose 3 numbers as the first digit (0 is excluded) and 3! possibilities for the other digits: 4!+2*3*3! = 60

ehild

Oh ! I made a mistake.. When I made case 1 and case 2 , I forgot to add the digit which I fixed at the end. Thanks for your efforts !