How can I solve questions like these???

[Curly213]

10 baseball teams plan to play each other 10 times in a certain season. How many total games will be played in the season?

Four different science books, six different art books and two different novels are to be arranged on a shelf. How many different arrangements are possible if (a) all the books of one type must stand together, and (b) only the science books must stand together?

answer
450
a) 4!6!2!3!
b) 4!9!

[hardikrs]

10 teams have to play every other team 10 times, so team 1 plays 90 games -- 10 with team 2,3,4,5,6,7,8,9
so every subsequent team has to play 10 less games as they have already played the previous team 10 times, for example 2 has to play only with team 3,4,5,6,7,8,9 since it already played 1 earlier.

so each time number of games reduced by 10..

90+80+70+60+50+40+30+20+10 = 450.

[JTRADER]

"Teams" problem, use permutations:

Total number of elements or "teams" = 10
Number of "teams" that play a game = 2
So, the formula would be:

Total Nº of elements! / [Nº of elements to combine! * (Total Nº of elem - Nº of elem to comb)!]

then:

10! / [2!*(10-2)!] = 10! / [2!*8!]

= (1*2*3*4*5*6*7*8*9*10) / (1*2*1*2*3*4*5*6*7*8)

Simplifying: (9*10) / (1*2) = 90 / 2 = 45

Since each team has to play not only once but 10 times with each other team, multiply the above output by 10;

45*10 = 450

[JTRADER]

"Books" problem:

For alternative a, use factorial for each set of books, to see in how many ways their elements can be ordered:

A set with n elements can be order in this way:

The first element can be ordered n time, the second n-1, the third n-2, etc...

So it would be: n*(n-1)*(n-2)......= n!

In the case of science books would be:
n= 4, then
4*(4-1)*(4-2)*(4-3) = 4*3*2*1 = 4!

Art books:
n=6 then the number of ways that can be order would be 6!

Novels:
n=2 then it would be 2!

Since the 3 different type of books can be taken as 3 different elements of a set (Books) you have 3! different ways to order them.

Then the number of ways that all books can be ordered is the product of the 4 above results:

4!6!2!3!

[tox4]

For Book b)

You have 4! ways to arrange 4 science books (SB) and there are 9! ways to arrange 8 other books and SB, eg. XXSBXXXXXX, hence the answer is 4!*9!

[antonydavis]

Originally Posted by JTRADER

"Teams" problem, use permutations:

Total number of elements or "teams" = 10
Number of "teams" that play a game = 2
So, the formula would be:

Total Nº of elements! / [Nº of elements to combine! * (Total Nº of elem - Nº of elem to comb)!]

then:

10! / [2!*(10-2)!] = 10! / [2!*8!]

= (1*2*3*4*5*6*7*8*9*10) / (1*2*1*2*3*4*5*6*7*8)

Simplifying: (9*10) / (1*2) = 90 / 2 = 45

Since each team has to play not only once but 10 times with each other team, multiply the above output by 10;

45*10 = 450

Isnt this a Combinations question? As order does not matter?