in how many ways can you lay out 8 red balls and 5 white balls in a row, so that the left most ball is red, and no two white balls are next to each other?
by saying r= red and w=white i have
rw.....rw......rw......rw.......rw with 3 red balls left over. the answers say to put the remaining 3 red balls in the 8 positions available, so the answer is 8 choose 3. But i do not get this because i dont see that there are 8 possible positions.
thanks for your help
2007-02-10 03:51:50 · 8 answers · asked by mond257 1 in Science & Mathematics ➔ Mathematics
ok. you can put:
all 3 red balls in the front,
2 in the front one in the back,
1 in the front 2 in the back or
all 3 in the back.
that's 4 right there.
now, you can put the other three so it is rrw...rrw...rrw...rw. there are 4 ways to do that.
rw...rrw...rrw...rrw
rrw..rw...rrw..rrw
rrw..rrw..rw..rrw
rrw...rrw...rrw...rw
there are the 8 ways. =)
2007-02-10 03:58:23 · answer #1 · answered by Ace 4 · 0⤊ 0⤋
Either the last ball is red or white.
If the last ball is red, you have 5 gaps in your row of 8 red balls, so they can be in this pattern
1,1,1,1,1,3 (think of commas as white balls)
There are 6 different places that could have 3 red balls.
Or they could be in this pattern:
1,1,1,1,2,2
There are 15 ways the two 2's could be arranged (6 choose 2)
Or a white ball could be on the right end and the red balls could be in these patterns
1,1,1,1,4 (commas are white balls again) - 5 possibilities
1,1,1,2,3 = 5*4 = 20 possibilities
1,1,2,2,2 = 10 possibilities
21 + 35 = 56 possibilities
2007-02-10 12:11:18 · answer #2 · answered by Steve A 7 · 0⤊ 0⤋
What you really have is this picture:
r...r...r...r...r...r...r...r...
The red balls have no *big* restrictions on where they show up, so we can have red balls right next to each other, etc - the only important one is that the first one must be red, which is why I don't have "..."s before the first red ball. The other 8 "..."'s represent the other spots where you can stick the 5 white balls. Because you only have 5 white balls, there are 8C5 different ways to place the white balls.
Answer: 8C5 = 8!/(5!3!) = (8*7*6)/(3*2*1) = 56
2007-02-10 12:00:09 · answer #3 · answered by mrfahrenbacher 3 · 0⤊ 0⤋
You actually need to look at it this way just watch and read my fellow Yahoo answeristic person!
r...r...r...r...r...r...r...r....
The red balls have no *big* restrictions on where they show up, so we can have red balls right next to each other, etc - the only important one is that the first one must be red, which is why I don't have "..."s before the first red ball. The other 8 "..."'s represent the other spots where you can stick the 5 white balls. Because you only have 5 white balls, there are 8C5 different ways to place the white balls.
2007-02-10 12:04:22 · answer #4 · answered by ☺C☺h☺a☺r☺l☺o☺t☺t☺e 3 · 0⤊ 0⤋
Collect all the balls up and place them in a box.
Then go have a cup of tea.
dont worry about it LOL
2007-02-10 12:08:10 · answer #5 · answered by PHIL D 2 · 0⤊ 0⤋
rrrrwrwrwrw
rwrrrrwrwrw
rwrwrrrrwrw
rwrwrwrrrrw
rrrwrrwrwrw
rrrwrwrrwrw
rrrwrwrwrrw
rrwrrwrrwrw
rrwrrwrwrrw
rwrrwrrwrrw
rwrwrrrwrrw
rwrwrrwrrrw
I think thats it with 12 possible positions.
2007-02-10 12:08:07 · answer #6 · answered by clever investor 3 · 0⤊ 0⤋
ur answer is: 8! * 5!=4838400
2007-02-10 12:04:29 · answer #7 · answered by Krish 5 · 0⤊ 0⤋
What would you wanna do that for?!
Buy a jigsaw puzzle
2007-02-10 11:56:23 · answer #8 · answered by ? 5 · 0⤊ 0⤋