Year 11 Permutations (3 U) (1 Viewer)

[hs17]

21 Numbers less than 4000 are formed from the digits 1, 3, 5, 8 and 9, without repetition.

a) How many such numbers are there?

b) How many of them are odd?

c) How many of them are divisible by 5?

d) How many of them are divisible by 3?

Can somebody please help me with part c and d

 

[vinlatte]

21 Numbers less than 4000 are formed from the digits 1, 3, 5, 8 and 9, without repetition.

a) How many such numbers are there?

b) How many of them are odd?

c) How many of them are divisible by 5?

d) How many of them are divisible by 3?

Can somebody please help me with part c and d

c) For the number to be divisible by 5, it must end in 5 or 0. So 5 is the last digit and must be fixed. Since the numbers have to be less than 4000, only 1 or 3 can be the first digit.
I think it would be 2 x 3!

d) I'm not really sure how to solve it either, but it might be helpful to know that any number that is a multiple of 3, you can add the digits and it will be a multiple of 3. Like 27 is 3 x 9, and the sum of the digits 2 + 7 is 9.

 

[Pedro123]

c) For this, you only include 4 digit numbers. The actual answer should be 2*3! (4 digit numbers) + 4*3 (3 digit numbers) + 4 (The 2 digit numbers) + 1 (The one-digit number)
d) I have to go to class, will post solution later. Sorry.

 

[Pedro123]

As stated, any number is divisible by 3 when the sum of its digits add to a number divisible by 3. This means adding a 3 or a 9 in any position in a number will not change whether it is divisible by 3.
The next insight is that 5+1 = 6 and 1+8 = 9, both of which are divisible by 3, but 5+8 is not divisible by 3. Let's begin by finding the 4 digit numbers. We know that the digits must either contain 3,9,1 and 5 or 3,9, 1 and 8. This gives us 2 combinations, each must contain a 3 or a 1 with the starting digit. This means there are 2*2*3! ways of arranging those numbers.
Now for a 3 digit number. We know this number can't include 3 AND 9, since then no matter what the last digit is, the number will not be divisible by 3. A similar argument shows we can't have 3 nor 9 in the number. Therefore, we must have 3 OR 9 in the number. This means that there are in total 4 combinations of digits - 1,5,3 + 1,6,9 + 1,5,9 + 1,6,3. 3! ways to get a combination of them means there are 4*3! 3 digit numbers.
Then 2 digit numbers. Obviously, our combinations are 1,5 + 1,8 + 3,9 which means we have 2*3 combinations.
There are finally only 2 1 digit combinations. Therefore the total number is:
2*2*3! + 4*3! + 2*3 + 2

 

[hs17]

The answer for part c is correct but part d seems to add up to a different number. I thought about it a bit more...so in the 4 digit combinations, you can't do 4! because the first number has to be 1 or 3 (question asks numbers less than 4000) so it would be (1 x 3!) x 2 -->just for working out the 4 digit part.

 

[Pedro123]

I updated my response look at it. The answer I got was twice your answer. Check if this is right.

 

[hs17]

Yes that’s right!

 

[Pedro123]

For future reference, the best ways (I find) for looking at combi problems in the HSC is to break up the problem. Most problems (Like the one above) lend themselves to divvying up the cases into groups which you can use the basic functions you learned in class. Another tip is looking for ways to group things the question isn't asking for i.e. if it asked for a group of all the odd numbers, you may consider instead finding all the even ones and subtracting that from the total number.
Solutions infrequently will come from one perfect, neat function - usually, it will be quite dirty work (Which is why personally, combinatorics is one of my least favourite math section, followed closely by geometry).

Good luck!

 

[hs17]

That's helpful advice, thanks!

 

[Directrix]

Just by looking, where was the question supplied from? Because there are some probability questions that could be thrown in my HY, at my teacher's discretion and this one is great for starting off.

Since I'm self learning permutations, due to online learning making me slack off, it seems more "light" compared to the other, algebraically and calc intensive topics to learn about, hence I believe this could be self-taught? Or am I just falling to the latter and am I just deluding myself?

 

[TheShy]

Also, where was the question supplied from? Since I'm self learning permutations, due to online learning making me slack off, it seems more "light" compared to the other, algebraically and calc intensive topics to learn about, hence I believe this could be self-taught? Or am I just falling to the latter and am I just deluding myself?

Nah perms and combs is quite easy self-learn as long as you have the answers to the questions. Hell, anything is self-learnable if you put in the time and effort. You can find a bunch of questions online and in pdfs.

 

[Directrix]

Great, wish me luck, I have made a Khan Academy account and I will go from there!

 

[hs17]

Just by looking, where was the question supplied from? Because there are some probability questions that could be thrown in my HY, at my teacher's discretion and this one is great for starting off.

Since I'm self learning permutations, due to online learning making me slack off, it seems more "light" compared to the other, algebraically and calc intensive topics to learn about, hence I believe this could be self-taught? Or am I just falling to the latter and am I just deluding myself?

This is from the Cambridge Yr 11 Ext 1 textbook