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# Kim has 10 identical lamps and 3 identical tables. How many ways are there for her to put all the lamps on the tables?

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Kim has 10 identical lamps and 3 identical tables. How many ways are there for her to put all the lamps on the tables?

Jun 19, 2018

#1
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Idk ?

Jun 19, 2018
edited by Guest  Jun 19, 2018
#2
+21931
0

deleted

Jun 20, 2018
edited by Guest  Jun 20, 2018
#9
+1846
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You shouldn’t delete your posts, Mr. BB.

According to experts, there should be:

[3 + 3 - 1] nCr 3 =5C3 = 5! / [5 - 3]!. 3! = 10 ways of putting the lamps on the tables.

You should just not make them in the first place!

You should also update your “expert” consultants list.  If you put my name at the top of the list,  then you will always have the correct answers.

GETSMART should really be at the top of your list though. He is your Christmas present and his genetic enhancement includes optimization to teach you. You are not a very good student, so he prefers to spend most of his time pecking bugs from your surroundings, while waiting for Thanksgiving.

GA

GingerAle  Jun 20, 2018
#10
+21931
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I have no idea who Mr BB is, WildChild.... I just deleted an incorrect answer I had posted as I was responding with a different (at least my second 'guess' was correct) answer to another poster.

G'Day!

ElectricPavlov  Jun 20, 2018
#11
+1846
+2

This is hilarious!

The reason I thought you were Mr. BB, is your presentation of this formula/equation

[3 + 3 - 1] nCr 3 =5C3 = 5! / [5 - 3]!. 3! = 10

in irregular mathematical syntax. I usually refer to this style as SLOP, because that is what it is. (It looks like the factorial of 3 = 10)

Mr. BB presents his formula/equations in a very similar sloppy format –though he often uses curly braces. Here’s an example: https://web2.0calc.com/questions/please-help_33396#r1

Mr. BB is a master at presenting slop, well beyond irregular mathematical syntax; he also has an incredibly high error rate—not just typos, it’s a bounty of misinformation presented with authority and irritation. This post elaborates and describes Mr. BB.

Excerpt:

...Mr. BB, the stubborn, relentless, intractable Blarney Banker of lore: a pseudo intellectual with a multiplicity of advanced dimwit degrees in arrogant stupìdity, a professor of misinformation, who teaches with authority and irritation.

Here’s a post he made to improve on one of your answers:

https://web2.0calc.com/questions/working-with-formulas-help

Here, demonstrating his finesse with Latin:

Here, demonstrating his finesse with coding. There’s also history about Mr. BB’s first troll, 14-YO Dragonlance. Along with other names, he called him the “Banker Man.” I call him the “Blarney Banker,” aka Mr. BB

https://web2.0calc.com/questions/a-question-of-rationality

Here’s an anonymous troll commenting on Mr. BB:

https://web2.0calc.com/questions/help-difficult#r5

I went blind laughing at this. I did again after rereading it for this post.

It’s so damn true!

There are many more if you care to peruse my profile –some are quite funny!

Anyway, EP, I probably wouldn’t have trolled you if your name were attached –at least not in this manner.

--

Curious question: do you ever use “G’day” as a valediction without the sarcastic overtones?

GA

GingerAle  Jun 20, 2018
#12
+21931
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Gothcha....   I didn't present the formula 'slop'..... that was another 'guest' poster....I just brute force listed all of the combos .....

(Sometimes I just answer a few Qs and don't get around to signing in to the forum.....then my name gets attached the next time I DO sign in to the forum)

G'day (again)!   (sometimes sarcastic....sometimes not)

ElectricPavlov  Jun 20, 2018
#3
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The formula says: P[n, k] =P[10, 3] =P[7,1] + P[7, 2] + P[7, 3] = 8 different ways as follows:

P stands for "Partitions", which means "How many ways can you make up a 10 out of 3 numbers from 1 to 10". So, I calculated it as being 8 ways: 1+1+8, 1+2+7, 1+3+6, 1+4+5, 2+2+6, 2+3+5, 2+4+4, 3+4+3. And that gives 8 ways of making a 10 out of 3 numbers from 1 to 10!!!!!!.

Note: 1+1+8, 1+8+1, 8+1+1 are considered all the same or just 1 combination.....etc.

P.S. Some expert should check this.

Jun 20, 2018
#4
+21931
0

Not sure what the answer is....but it has to be more than 8 doesn't it?:

10   0    0

9     1    0

8      2   0

7       3  0

6       4   0

5       5   0

8     1     1

7     2     1

6     2     2

6      3    1

5       4   1

5     3     2

4     3    3

4     4    2

fourteen ways?

ElectricPavlov  Jun 20, 2018
edited by Guest  Jun 20, 2018
#7
+1846
+2

OK. I’m an expert.

The formula says: P[n, k] =P[10, 3] =P[7,1] + P[7, 2] + P[7, 3] = 8 different ways as follows:

The formula is correct, but it is incomplete for this question.

P stands for "Partitions", which means "How many ways can you make up a 10 out of 3 numbers from 1 to 10". So, I calculated it as being 8 ways: 1+1+8, 1+2+7, 1+3+6, 1+4+5, 2+2+6, 2+3+5, 2+4+4, 3+4+3. And that gives 8 ways of making a 10 out of 3 numbers from 1 to 10!!!!!!.

This is the correct interpretation for the incomplete formula.

Note: 1+1+8, 1+8+1, 8+1+1 are considered all the same or just 1 combination.....etc.

There are no combinations here, only partitions.

No banana for you.

GA

GingerAle  Jun 20, 2018
#5
0

I think "zeros" are NOT allowed. You have to make a 10 in groups of 3 numbers from 1 to 10.

See the formula used here, under " identical and identical":

https://www.careerbless.com/aptitude/qa/permutations_combinations_imp7.php

Jun 20, 2018
#8
+1846
+1

A partition set for a number does not normally have empty sets, but a table can have zero lamps, so by implication there is a zero.

There are three sets for this question, and the question allows for empty sets.

GA

GingerAle  Jun 20, 2018
#6
+1846
+3

The general formula:

$$\large \sum\limits_{i=1}^{T}P(L, i) \tiny \leftarrow \text{P is the partition(s) of i in sets of L; T = # of tables, L = # of lamps, and i is the index of T}\\\\ \large \sum\limits_{i=1}^{10}P(3, i) = 14 \tiny \leftarrow \text{With specified parameters. }\\$$

Descriptive list:

10 on one table & none on the others

9 on one table & 1 on another & none on the other

8 on one table & 2 on another & none on the other

7 on one table & 3 on another & none on the other

6 on one table & 4 on another & none on the other

5 on one table & 5 on another & none on the other

8 on one table & 1 on another & 1 on the other

7 on one table & 2 on another & 1 on the other

6 on one table & 3 on another & 1 on the other

6 on one table & 2 on another & 2 on the other

5 on one table & 4 on another & 1 on the other

5 on one table & 3 on another & 2 on the other

4 on one table & 4 on another & 2 on the other

4 on one table & 3 on another & 3 on the other

GA

Jun 20, 2018
#13
+48
0

i think there are 14 ways

Jun 20, 2018
#14
+48
0

oops didn't see the post before that

Jun 20, 2018