Question

11 student are in a waitlist for a course. two of the students in the list are named ed and ned. how many different ways are there for the students to be ordered in the list so that ed is higher in the list than ned?

Answer 1

The different ways that are there for the students to be ordered in the list so that Ed is higher on the list than Ned is 19,958,400

In the order, if Ned comes first on the list then Ed will not be on the list before Ned. (remember for Ed to be higher than Ned on the list, it should come after Ned on the list).

So if Ned is first on the list, then all the remaining (including Ed) can occur anywhere in the remaining list. Hence the total permutations will be 10!

Ned's corresponding choices: 1 109 8 7 6 5 4 3 -2 1

If Ned comes in the second position, then Ed can't come in the first position. Hence in the first position, there are only 9 choices.

*(excluding Ed) Ned * corresponding choices:

= 9*9!

*(including Ed (including Ed) Ned corresponding choices: *k

= 9*8*8!

Total combinations will be:

10! + 9*9! + 9*8*8! + 9*8*7*7! + 9*8*7*6*6! + 9*8*7*6*5*5! + 9*8*7*6*5*4*4! + 9*8*7*6*5*4*3*3! +9*8*7*6*5*4*3*2*2! + 9*8*7*6*5*4*3*2*1*1! .

= 19,958,400

We can directly compute it. Since there will be a total of 11! combinations in which 11!/2 times Ed comes after Ned and 11!/2 times Ed comes before Ned.

and 11!/2 = 39,916,800/2 = 19,958,400

To learn more about permutation and combination,

brainly.com/question/1216161

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