Q:permutation (1 Viewer)

[freaking_out]

i have got two questions:


1) 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?

In this question there was also a part asking how many are divisible by 5 and i answered that by the fact that anything divible by 5 will end with 0 or 5 (5 in this case) and hence i found the permutations of numbers ending with 5. but with divisibility of 3 i cannot do it like that since no. divisible by 3 can end with 1,9,3,6 etc.

2) In how many ways can a boat crew of eight women be arranged if three of the women can only row on the bow side and two others can only row on the stroke side?

Thanx in advance and i have got the answer but just need the method.

 

[kini mini]

Originally posted by freaking_out



. but with divisibility of 3 i cannot do it like that since no. divisible by 3 can end with 1,9,3,6 etc.

I can't remember how to do the first one off the top of my head unfortunately, are there any good divisibility tests for 3? Must be some cunning trick here but it hasn't occured to me yet.

2) In how many ways can a boat crew of eight women be arranged if three of the women can only row on the bow side and two others can only row on the stroke side?

Thanx in advance and i have got the answer but just need the method.

I assume it's four women on each side, I don't know what you mean by bow and stroke though . Isn't it usually port and starboard? I've never done any rowing myself.

Arrange 3 on bow x Arr 2 on stroke x Arrange last 3 in 3 places

So 4.3.2 x 4.3 x 3.2.1

 

[spice girl]

Originally posted by freaking_out
i have got two questions:


1) 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?

In this question there was also a part asking how many are divisible by 5 and i answered that by the fact that anything divible by 5 will end with 0 or 5 (5 in this case) and hence i found the permutations of numbers ending with 5. but with divisibility of 3 i cannot do it like that since no. divisible by 3 can end with 1,9,3,6 etc.

2) In how many ways can a boat crew of eight women be arranged if three of the women can only row on the bow side and two others can only row on the stroke side?

Thanx in advance and i have got the answer but just need the method.

1) For divisibility test for 3 is that the sum of its digits must be divisible by 3. It doesn't matter which order the digits are in.

Also, 1 gives remainder 1, and 5, 8 both give remainder 2 when divided by 3.

Case1: single figures
3, 9
# = 2

Case2: double figures
39, 93
# = 2

Case 3: three figures
since only 2 digits divisible by 3, thus at least 1 digit not divisible by 3. Thus we need at least 2 digits not divisible by 3. These pairs of digits are (1, 5) and (1, 8). Note that the 3 digits (1,5,8) does not work.
so: (1,3,5) # = 6
(1,5,9) # = 6
(1,3,8) # = 6
(1,8,9) # = 6
total # = 24

Case 4: four figures
must omit one digit out of possible 5. This digit is either 5 or 8 (same reasoning as given above). Note that the first digit is either 1, 3, to make the number < 4000.
(omit 5) # = 2 * 6 = 12
(omit 8) # = 2 * 6 = 12
total # = 24

So total in all cases is 2+2+24+24.

2) General method is to accomodate the troublesome women first:
Place the 3 bow-women: choices = 4 * 3 * 2 (since there are four seats to place the first woman, etc, etc)
Place the 2 stroke women (don't you mean starboard?): choices = 4*3.

We have 3 seats left for the normal women. choices = 3 * 2 * 1

Total # = 4 * 3 * 2 * 4 * 3 * 3 * 2 = 1728

 

[kini mini]

Re: Re: Qermutation

Originally posted by spice girl


2) General method is to accomodate the troublesome women first:
Place the 3 bow-women: choices = 4 * 3 * 2 (since there are four seats to place the first woman, etc, etc)
Place the 2 stroke women (don't you mean starboard?): choices = 4*3.

We have 3 seats left for the normal women. choices = 3 * 2 * 1

Total # = 4 * 3 * 2 * 4 * 3 * 3 * 2 = 1728

Thought that was it, nice one

Hmmm, I dunno if I would say "troublesome women" in this new age of political correctness . How about special cases?

 

[freaking_out]

yeah the answer for the second question is 1728 (you know what i didn't know what bow and stroke meant either thats why i probably couldn't solve it ) and with the "troublesome woman" ... ah you must be part of the feminist movement or somefin'

with the second question, the answer is 50 so its either the t.book is wrong or the solution prob. over counted by 2 (damn those exceptions!)

at the moment i just looked at the answers and tommorow i will look at the solution... coz now i can't think in this new year atmosphere .

 

[freaking_out]

Place the 3 bow-women: choices = 4 * 3 * 2 (since there are four seats to place the first woman, etc, etc)

actually i couldn't resist looking at the solutions!!
yeah now i get where i went wrong coz when i tried to do it i didn't know what bow meant so when accommodating the 3 bow woman i said 3*2*1... ... alright i am not a rower, you got me there!!!

wif the second question and that, i wanna know how i can prepare my self for all this "divisible by x" type of questions coz obvioulsy i didn't know anythin' about this concept.

 

[freaking_out]

Case2: double figures
39, 93

about case 2, i wanna know how you got these no. and also what about numbers like 51, 15, 81, 18 etc.

since only 2 digits divisible by 3, thus at least 1 digit not divisible by 3. Thus we need at least 2 digits not divisible by 3. These pairs of digits are (1, 5) and (1, 8). Note that the 3 digits (1,5,8) does not work.

can you clarify what this means? coz i get the first bit about how there must be atleast one digit in the 3 digit no. that is not divisible by 3 (since there is only 2 digits that are divisible by 3 available)... after that i don't get the bit after that.

 

[spice girl]

sorry, u're right with 15, 18, 51, 81. Silly mistake number 17365 for me.

As for case 3, there are only 3 cases: either 1, 2, or 3 digits out of the three, that are not divisible by 3, 1 digit definitely won't give us what we want cos if we only had one digit not divisible by 3, the whole number will not be divisible by 3 (for obvious reasons). Neither will 3 digits not divisible by 3 (there's only one way that can happen: 1,5,8) give us a number divisible by 3. Thus, 2 digits is our only option.

 

[freaking_out]

Silly mistake number 17365 for me

well if thats a figure for a 99% maths student, then my one prob. have to be 4 times that amount.

anyway with question 1 is the textbook answer 50 correct? or is it now more than 52 (since you didn't count51,18 etc.)

 

[spice girl]

seems like 56 now...

 

[freaking_out]

damn the bloody textbook for putting me off!!

 

[drolle]

Originally posted by freaking_out


well if thats a figure for a 99% maths student, then my one prob. have to be 4 times that amount.

Didn't you know? The true measure of someone's mathematical ability is the number of mistakes they make -- the more mistakes, the better they are

 

[freaking_out]

Originally posted by drolle


Didn't you know? The true measure of someone's mathematical ability is the number of mistakes they make -- the more mistakes, the better they are

ok fine then my amount is a quarter of that amount spice girl has...geez