Question 1156747: How many ways can we arrange 5 boys and 10 girls in a line where all the 5 boys are always together?
Found 3 solutions by greenestamps, jim_thompson5910, ikleyn:
Answer by greenestamps(13200)   (Show Source):
You can put this solution on YOUR website!
Put the 5 boys together in a group; within that group they can be arranged in 5!=120 different ways.
Put the 10 girls in a line. There are 10! ways they can be arranged.
There are 11 different places in that line (including both ends) where the group of boys can be.
The total number of arrangements with all the boys together is (5!)(10!)(11).
Answer by jim_thompson5910(35256)  (Show Source):
You can put this solution on YOUR website!
Answer: 4,790,016,000
This number is a little over 4.79 billion
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How to get this answer:
We want the boys to stay together, so we'll consider all the boys to be one "person". Subtract off the 5 boys to have 15-5 = 10 people left (just the girls for now). Then add in that "person" to get 10+1 = 11 people total. If you want, you can think of the 11 refering to slots. Ten slots will have one person each (one girl each) while the eleventh slot really has five boys in it.
There are 11! = 11*10*9*8*7*6*5*4*3*2*1 = 39,916,800 ways to arrange these 11 slots around. The exclamation mark is factorial notation.
Within any given permutation mentioned above, there are 5! = 5*4*3*2*1 = 120 ways to arrange the five boys in the block they stay in.
So overall there are 120*39,916,800 = 4,790,016,000 ways to arrange all the students such that the five boys stick together.
Answer by ikleyn(52802)  (Show Source):
You can put this solution on YOUR website! .
5! = 120 ways to arrange the boys in their group of 5 boys.
11 ways to place this group as one single unit among 10 girls,
and 10! ways to order the girls.
In all, there are 5! * 11 * 10! ways.
You can calculate the value on your own.
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