The number of ways in which four particular persons A,B,C,D
and six more persons can stand in a queue so that A always
stands before B, B before C, and C before D, is ?

- mathmath333

- chestercat

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- mathmath333

\(\large \color{black}{\begin{align}
& \normalsize \text{ The number of ways in which four particular persons A,B,C,D }\hspace{.33em}\\~\\
& \normalsize \text{ and six more persons can stand in a queue so that A always }\hspace{.33em}\\~\\
& \normalsize \text{ stands before B, B before C, and C before D, is ? }\hspace{.33em}\\~\\
\end{align}}\)

- dan815

|dw:1442643874411:dw|

- mathmath333

this is one of the case

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- mathmath333

|dw:1442644105548:dw|

- dan815

10!/4!

- dan815

there are 10! ways to arrange these 10 persons
now for each of those arrangements
there is some arrangement of ABCD 4! ways of them, we only want the arrangement A B C D so 1 for every one of thsoe 4! ways
thus 10!/4!

- mathmath333

but the ABCD can be in many ways like this
|dw:1442644275123:dw|

- dan815

do you want clearer explaination

- dan815

for example lets take some random arrangement in 10!
1 2 3 4 A 5 B 6 C D
okay for this arrangement there are 4! other ways we count when A B C D can be placed in way , and the numbers 1 to 6 are left in same spot

- mathmath333

is this (below) a valid set up|dw:1442644423498:dw|

- dan815

we count that in the 10!

- dan815

so taking 1 for every 4! we will take only the A B C D case for each of those arrangements in 10!

- mathmath333

ok

- dan815

you have any questions about this?

- mathmath333

let me read and think

- dan815

ill make it clearer okay

- dan815

for 10! ways we will have the arrangement of every single possible way of placing these 10 people

- dan815

let us look at some of these arragenment sequences

- dan815

|dw:1442644766778:dw|

- dan815

there are 4! ways for this one arrangement of how you play these 6 people in the 10 spots

- dan815

out of these 4! ways , we only want one of them where its A B C D

- dan815

so another way less intuitive would now be this means all we care about is how we place the 6 people in the 10 spots

- mathmath333

ok i got it

- dan815

ok

- mathmath333

|dw:1442645081498:dw|

- dan815

welcome!

Looking for something else?

Not the answer you are looking for? Search for more explanations.