duttsit wrote:
vikramm wrote:
rchadha wrote:
OA is 6*5!
Anyone knows a simple methos to arrive at that?
Actually, its fairly simple ->
Identical red balls can be arranged in only 1 set way. That is a gap between each of them.
The 5 different green balls amongst themselves can be arranged in 5! ways and
6 ways between the gaps created by Red as shown by bsmith75 which is missing one combo (see bold).
GGRGRGRGR = 5! combos
GRGGRGRGR = 5!
GRGRGGRGR = 5!
GRGRGRGGR = 5!
GRGRGRGRG = 5!
RGRGRGRGG I.e 6*5!*1 = 6*5!
How did you get 6 ways to put 4 identical Red balls.
We have 6 positions available:
X G X G X G X G X G X
in these 6 position we need to put 4 Red balls (order not important)
wouldn't it be: 6C4 = 10
for example, why did you exclude these arrangements:
RGRGRGGRG
RGRGRGGGR
RGRGGRGRG
Thanks duttsit, You are correct... my bad...
The combinations you list are valid...
Trying to avoid matching the answers... but just extending your method...
Why do we need 6C4? The four balls are identical.. shouldn't it be 6C1?
If we remove the Green Balls in the mix scenario... 4 identical Red balls together can be arranged in only one possible way.. as they are identical. Right? So, going back to -> X G X G X G X G X G X -> It doesnt matter which Red Ball goes where (as long as it is not next to each other) which means 6C1 ways... am I missing anything?
_________________
-Vikram
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