Question

Find the number of ways in which three numbers can be selected from the set \(\{1,2,3,\ldots,4n\}\), such that the sum of the three selected numbers is divisible by 4.

Answer 1

1+2+3=4n is this ?

Answer 2

type.

Answer 3

anyway whatever :p

Answer 4

n numbers divisible by 4, nC3 <--- perhaps this will be for all three numbers divisible by 4


on the other hand, we we choose odd number, say 1, then we have to choose 3 or 7, 5 won't work, I think there are n such numbers, n*(n-1)*n

so total wouldbe nC3 + n*(n-1)*n <--- just a guess

Answer 5

How about we express every number as \(4n,4n+1,4n+2\) and \(4n+3\) ?
Case1. 4n, 4n, 4n
Case2. 4n, 4n + 1, 4n + 3
Case3. 4n + 2, 4n + 2, 4n
Case4. 4n+1,4n+1,4n+2

Would this work?

Answer 6

hey man, can you choose the same no twice??

Answer 7

No? it's just the form of number, not the number itself.

Answer 8

Case4. 4n+1,4n+1,4n+2 <-- not sure about this

Answer 9

looks like i missed case 3

Answer 10

Since there are 4n numbers, each form must have n numbers.

For Case1. n*(n-1)*(n-2)
Case2. n*n*n
Case3. n*(n-1)*n
Case4. n*(n-1)*n

Answer 11

is it combination or arrangement??

Answer 12

experimentx just the form of number. For example \(\underbrace{1}_{4\cdot0 +1} + \underbrace{5}_{4\cdot1 + 1} + \underbrace{2}_{4\cdot 0 +2} = 8\).

Answer 13

Hmm so total number of ways must be n^3 + n(n-1)(n-2) + 2(n)*(n-1)*n

Answer 14

seems all right to me.

Answer 15

But I can't be sure, I don't have the answer. :(

Answer 16

For Case1. n*(n-1)*(n-2) <--- same pattern so it should be decreasing
Case2. n*n*n <--- three different pattern
Case3. n*(n-1)*n <---- two same pattern one different pattern
Case4. n*(n-1)*n <--- two same pattern one different patter

Answer 17

if it is arrangement, then it should be that way.
if it's just selection, then we have to divide it by 3!

Answer 18

ohh yeah, it's selection. so divide it by 3!.

Answer 19

i think so.

Answer 20

hmm \[C_1\cdot C_1 \cdot C_1 + 2\cdot C_2\cdot C_1 +C_3\]Is this right? nCr -> Cr.

Answer 21

not really,
1) nC3
2) nC1 x nC1 x nC1
3) nC2 x nC1
4) nC2 x nC1

I think it would be this way, for all.

Answer 22

Oo .. I come to think now, dividing by 3! was only in the first case.
I would have been incorrect.

Answer 23

hmm thanks :-)

Answer 24

well ... most of the job you did yourself.
still ... you are welcome.