perms and combs Q (1 Viewer)

[Makematics]

Hi guys, here is a perms and combs question that i am stuck on... it would be great if you could help me out.
Numbers less than 4000 are formed from the digits 1,3,5,8 and 9, without repetition.
How many of them are divisible by 3?
i get 56 as my answer, but the textbook says it is 50...

 

[RealiseNothing]

I get 56 too.

 

[cineti970128]

really? i ve got 50
anyone?

 

[cineti970128]

the working is :

firstly you have to know that any number is divisble by 3 if all digits sum to be divisble by 3
ie (12132) is divisble by 3 because 1+2+1+3+2 = 9 divisble by 3.

Now by T AND E for 4 digit values (3,1,8,9) , and (3,5,1,9) their sums are divisble by 3
the numbers are less than 4000 hence start with 3 or 1
hence 2x3! + 2x3! = 24

Now BY T AND E for 3 digit values (1,8,9), (1,5,9) and (3,1,5)
ie 3x3! = 18

Now BY T AND E for 2 digit (3,9) (8,1) and (5,1)
ie 3x2! = 6

and 3, 9 ie 2

hence 24+18+6+2 = 50


Makematics could you show me how you got 56 maybe i missed out on something

 

[Makematics]

the working is :

firstly you have to know that any number is divisble by 3 if all digits sum to be divisble by 3
ie (12132) is divisble by 3 because 1+2+1+3+2 = 9 divisble by 3.

Now by T AND E for 4 digit values (3,1,8,9) , and (3,5,1,9) their sums are divisble by 3
the numbers are less than 4000 hence start with 3 or 1
hence 2x3! + 2x3! = 24

Now BY T AND E for 3 digit values (1,8,9), (1,5,9) and (3,1,5)
ie 3x3! = 18

Now BY T AND E for 2 digit (3,9) (8,1) and (5,1)
ie 3x2! = 6

and 3, 9 ie 2

hence 24+18+6+2 = 50


Makematics could you show me how you got 56 maybe i missed out on something

hey man, thanks for trying it out! i used the same method as you. i think you forgot the 3 digit combination of (1,3,8) which adds on another 3! on to the answer.

 

[cineti970128]

hey man, thanks for trying it out! i used the same method as you. i think you forgot the 3 digit combination of (1,3,8) which adds on another 3! on to the answer.

OH... i did not see that.. in that case i think you re right 56 it is.

 

[Makematics]

here's another one, should be fairly easy, but i can't get it... might even be 2u probability idk lol.
Cards are drawn one at a time, without replacement, from an ordinary pack of 52 cards. Cards drawn have the following values: Aces score 1 point, tens, jacks, queens and kings score 10 points, and cards from two to nine score as many points as the numbers they carry (i.e. twos score 2 points, etc.) Find the probability that
(a) the first card drawn scores at least 9 points. (answer is 5/13) got this one
(b) the first two cards drawn each score at least 9 points (answer is 1/13)
(c) the first two cards drawn score at least 18 points altogether (answer is 8/15)
(d) the first two cards drawn each score at least 9 points given that they score at least 18 points altogether. (answer is 359/1170)

 

[RealiseNothing]

b)

???

 

[Makematics]

b)

???

that's exactly what i did, apparently its wrong :S

 

[braintic]

b)

???

Your answer is correct. And even if their answers to (b) and (c) were correct, their answer to (d) could not be. Since (c) contains all of (b), Bayes' Theorem (not in the course) says that (d) is 1/13 divided by 8/15, which is not their answer.

 

[Makematics]

another q. A three-digit number is formed from the digits 3,4,5,6 and 7 (no repetitions allowed). Find the probability that the number contains the digits 3 or 5.
I think i'm getting confused by the wording, can't get the answer which is 9/10.

 

[braintic]

another q. A three-digit number is formed from the digits 3,4,5,6 and 7 (no repetitions allowed). Find the probability that the number contains the digits 3 or 5.
I think i'm getting confused by the wording, can't get the answer which is 9/10.

Even though technically order is important, it is a probability question in which order plays no special role, so as long as the denominator is counted the same way, we need not worry about order.
Contains 3: 4C2=6 (Choose 2 numbers from 4 to go with the 3)
Contains 5: Also 6
Contains 3 & 5: 3C1=3 (Choose 1 number from 3 to go with the 3 and 5)
Total combinations: 5C3=10

Probability = (6 + 6 - 3) / 10 = 9/10 (Subtracting to avoid double counting)

OR (EASIER)

Only combination without 3 or 5 is 467.
So 9/10 combinations have a 3 or 5.

 

[RealiseNothing]

Even though technically order is important, it is a probability question in which order plays no special role, so as long as the denominator is counted the same way, we need not worry about order.
Contains 3: 4C2=6 (Choose 2 numbers from 4 to go with the 3)
Contains 5: Also 6
Contains 3 & 5: 3C1=3 (Choose 1 number from 3 to go with the 3 and 5)
Total combinations: 5C3=10

Probability = (6 + 6 - 3) / 10 = 9/10 (Subtracting to avoid double counting)

OR (EASIER)

Only combination without 3 or 5 is 467.
So 9/10 combinations have a 3 or 5.

That's such a teaser of a solution isn't it.

 

[Makematics]

very clever! thanks for your help! i think i interpreted the question as either 3 or 5, but not both.

 

[Makematics]

sorry for the spam but i'm not the greatest at perms and combs

Find how many ways the integers 1,2,3,4,5,6,7,8 can be placed in a circle if at least three odd numbers are together.

i tried finding the number of ways the integers can be placed using the two cases of the four odd numbers being together and three odd numbers being together, but i don't seem to be getting the answer. (answer is 2304)

 

[braintic]

sorry for the spam but i'm not the greatest at perms and combs

Find how many ways the integers 1,2,3,4,5,6,7,8 can be placed in a circle if at least three odd numbers are together.

i tried finding the number of ways the integers can be placed using the two cases of the four odd numbers being together and three odd numbers being together, but i don't seem to be getting the answer. (answer is 2304)

Case I: Exactly 3 odds together

Pick the three odd numbers you want together from the 4 available: 4 ways
Arrange them: 3! ways
Place the other odd number in any of the non-adjacent positions: 3 ways
Place the evens in the other 4 slots: 4! ways

4 times 3! times 3 times 4! = 1728

Case II: 4 odds together

Arrange the 4 odds: 4! ways
Place the 4 evens in the remaining slots: 4! ways

4! times 4! = 576

1728+576 = 2304

 

[Makematics]

Case I: Exactly 3 odds together

Pick the three odd numbers you want together from the 4 available: 4 ways
Arrange them: 3! ways
Place the other odd number in any of the non-adjacent positions: 3 ways
Place the evens in the other 4 slots: 4! ways

4 times 3! times 3 times 4! = 1728

Case II: 4 odds together

Arrange the 4 odds: 4! ways
Place the 4 evens in the remaining slots: 4! ways

4! times 4! = 576

1728+576 = 2304

thank you for that, I just forgot to multiply by the four that is highlighted. are there four ways to choose three odd numbers because it is 4C3, or something else?

 

[braintic]

thank you for that, I just forgot to multiply by the four that is highlighted. are there four ways to choose three odd numbers because it is 4C3, or something else?

4C3 if you like. But I think its easier to think of leaving one behind, and there are 4 choices for the one to leave out.