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Maths question(difficult)


Mllcg
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1st Permutation...

 

There are 4 Variants.

Each Variant can either WIN or LOSE, for a total of 3 times.

... If a Variant WINS, the value MUST range from 2-3.

If a Variant LOSES, the value MUST range from 0-1.

The TOTAL number of WINS & LOSSES for all 4 Variants be add up to 6 WINS & 6 LOSSES.

 

 

 

2nd Permutation...

Each Variant's TOTAL Value MUST range from 0-9.

All 4 Variants MUST ADD up to 18.

Each Variant's TOTAL Value can share the same number with another Variant.

 

 

I'm looking for all POSSIBLE answers that will lead to either of the following...

 

- A permutation that leads to 3 Variants having the same number

- A permutation that leads to 2 Variants having the same number

- A permutation that leads to 2 Variants having the same number, but their value is in between 2 other Variants' value.

- A permutation that leads to 2 Variants having the same number, and their value is lesser than the other 2 Variants.

 

 

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1st Permutation...

 

There are 4 Variants.

Each Variant can either WIN or LOSE, for a total of 3 times.

... If a Variant WINS, the value MUST range from 2-3.

If a Variant LOSES, the value MUST range from 0-1.

The TOTAL number of WINS & LOSSES for all 4 Variants be add up to 6 WINS & 6 LOSSES.

 

 

 

2nd Permutation...

Each Variant's TOTAL Value MUST range from 0-9.

All 4 Variants MUST ADD up to 18.

Each Variant's TOTAL Value can share the same number with another Variant.

 

 

I'm looking for all POSSIBLE answers that will lead to either of the following...

 

- A permutation that leads to 3 Variants having the same number

- A permutation that leads to 2 Variants having the same number

- A permutation that leads to 2 Variants having the same number, but their value is in between 2 other Variants' value.

- A permutation that leads to 2 Variants having the same number, and their value is lesser than the other 2 Variants.

 

 

pardon me but Kum Pui?

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Neutral Newbie

1st Permutation...

 

There are 4 Variants.

Each Variant can either WIN or LOSE, for a total of 3 times.

... If a Variant WINS, the value MUST range from 2-3.

If a Variant LOSES, the value MUST range from 0-1.

The TOTAL number of WINS & LOSSES for all 4 Variants be add up to 6 WINS & 6 LOSSES.

 

 

 

2nd Permutation...

Each Variant's TOTAL Value MUST range from 0-9.

All 4 Variants MUST ADD up to 18.

Each Variant's TOTAL Value can share the same number with another Variant.

 

 

I'm looking for all POSSIBLE answers that will lead to either of the following...

 

- A permutation that leads to 3 Variants having the same number

- A permutation that leads to 2 Variants having the same number

- A permutation that leads to 2 Variants having the same number, but their value is in between 2 other Variants' value.

- A permutation that leads to 2 Variants having the same number, and their value is lesser than the other 2 Variants.

 

I suppose your questions only refer to the 2nd permutation?

 

- A permutation that leads to 3 Variants having the same number

Using trial-and-error:

0 0 0 18 (rejected)

1 1 1 17 (rejected)

2 2 2 12 (rejected)

3 3 3 9

4 4 4 6

5 5 5 3

6 6 6 0

4 different cases, with 4 combinations (4C3) for each.

4 x 4 = 16 cases

 

- A permutation that leads to 2 Variants having the same number (never state if its single or double pair of 2 Variants)

Using trial-and-error again:

Single pair of 2 Variants having the same number

1 1 9 7

2 2 9 5

2 2 8 6

3 3 8 4

3 3 7 5

4 4 9 1

4 4 8 2

4 4 7 3

5 5 8 0

5 5 7 1

5 5 6 2

11 different cases, with 12 combinations (4!/2!) for each.

11 x 12 = 132 cases

Double pair of 2 Variants having the same number

0 0 9 9

1 1 8 8

2 2 7 7

3 3 6 6

4 4 5 5

5 different cases, with 6 combinations (4!/(2! x 2!)) for each.

5 x 6 = 30 cases

Total cases= 132 + 30 = 162 cases

 

- A permutation that leads to 2 Variants having the same number, but their value is in between 2 other Variants' value.

Eliminating from the previous question,

4 4 9 1

4 4 8 2

4 4 7 3

5 5 8 0

5 5 7 1

5 5 6 2

6 different cases, with 12 combinations (4!/2!) for each.

6 x 12 = 72 cases

 

- A permutation that leads to 2 Variants having the same number, and their value is lesser than the other 2 Variants (never state if other 2 Variants can be the same).

Eliminating from the previous 2 questions,

1 1 9 7

2 2 9 5

2 2 8 6

3 3 8 4

3 3 7 5

5 different cases, with 12 combinations (4!/2!) for each.

5 x 12 = 60 cases

0 0 9 9

1 1 8 8

2 2 7 7

3 3 6 6

4 4 5 5

5 different cases, with 6 combinations (4!/(2! x 2!)) for each.

5 x 6 = 30 cases

Total= 60 + 30 = 90 cases

 

Haha my solutions taken with a pinch of salt [sly]

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Neutral Newbie

ya.

 

but cheem still

 

I think shld have some formula...

trial and error only for desperate measures:)

[where often many cases may be missed out]

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Neutral Newbie

What's 6 divided by 2 (1+2)?

 

Answer 1 or 9?

 

If 6 divided by 2(1+2), it is 1.

If 6 divided by 2, then multiply by (1+2), it is 9.

The space btw 2 (1+2) is unclear:)

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What's 6 divided by 2 (1+2)?

 

Answer 1 or 9?

 

 

9.

 

 

using bodmas rule..

 

x or / will supercede + & - .

 

so 6 / 2(1+2) = left divide 1st..

 

3 ( 1+2) = 9. my 2 cents.. :D

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1st Permutation...

 

There are 4 Variants.

Each Variant can either WIN or LOSE, for a total of 3 times.

... If a Variant WINS, the value MUST range from 2-3.

If a Variant LOSES, the value MUST range from 0-1.

The TOTAL number of WINS & LOSSES for all 4 Variants be add up to 6 WINS & 6 LOSSES.

 

 

 

2nd Permutation...

Each Variant's TOTAL Value MUST range from 0-9.

All 4 Variants MUST ADD up to 18.

Each Variant's TOTAL Value can share the same number with another Variant.

 

 

I'm looking for all POSSIBLE answers that will lead to either of the following...

 

- A permutation that leads to 3 Variants having the same number

- A permutation that leads to 2 Variants having the same number

- A permutation that leads to 2 Variants having the same number, but their value is in between 2 other Variants' value.

- A permutation that leads to 2 Variants having the same number, and their value is lesser than the other 2 Variants.

 

 

return to teacher liao.. i sux at statisitcs..

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(edited)

Wah kaoz......catch no balls. [crazy]

 

The only permutation i know is Spools 4D.

 

$24 for 4 different number

$12 for 2 diff number henceforth..........haha [laugh]

Edited by Hiphiphoray
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