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There are 8 letter A,B,C,D,E,F,G,H
In a table in how many ways can the 8 letters be arranged such that
A,B,C are only placed in column 1 and E,F are placed only in column 2.
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(and leave the last slot in column 1 empty)
yes, I would 4 places to put A, 3 to put B, 2 to put C
next , we fill the last slot , but we cant use A,B,C , E , F
that leaves D,G,H
so 4*3*2*3 to fill the first column
4*3 to fill the 2nd column with E and F
that leaves 2 slots and 2 letters
so all together
4*3*2*3 * 4*3*2
why didnt u add like 4*3*2*3+4*3*2
we have 4*3*2*3 = 72 ways to fill the first column 1
say we had only 1 slot for column2 and we could put in A , B or C
if we put in A in the second column, then we would have 72 patterns
if we put in B we have the same 72 patterns but with a B , so different from the first set
ditto for C
in other words, if we have 6 patterns in the first column and 6 in the 2nd,
we multiply to get the number of different patterns.
4*3*2 for abc
4*3 for ef
then 3! for the remaining