Permutations

Question

Comments

It will be $\frac{8!}{2} = 20160$.

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20160= (1+2+3+4+5+6+7)$\times$(6!)

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why did you divide it by 2 @jason

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@shivanisrivarshini because in all possible permutations half permutation are there in which b followed by d and half permutation are there in which d followed by b.

You can take the example of $\{1,2,3\}$ having 3 permutations in which 1 followed by 2 and 3 in which 2 followed by 1.

Answer 1

Few ways to do this First is the number of permutation where b appear before d, is same as the number of permutation where d appear before a d= total number of permutation /2 as half will have the first case and another half will have the second case.

so $\frac{8!}{2}$ one can try for 3 or 4 numbers and see

Method 2

we have 8 positions we fix d in the first position then we will have 7 positions left for b, and rest all can be arranged in 6! ways

fix d in the second position we have 6 positions left for b, rest in 6! ways

1st = 6!*7

2nd = 6!*6

7th = 6!*1

adding all 6![7+6+5+4+3+2+1] = 6!*$\frac{7*8}{2}$ = $\frac{8!}{2}$

Comments

I am not getting your method TESLA pls give some example for this

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Take ABC and find the number of strings.in which A comes before B it will be 3!/2

Take ABCD and find same A before B it will be 4!/2