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METHOD I:Originally posted by 1234567
ok that's from the specimen paper
the model answer goes like this:
total num of diff selection = 8c3 = 56
num of sleections which include both e's = 6
num of selections which dunt include both e's = 50
num of arrangement = 50 x 3! + 6x 3!/2! = 318
which i dunt undertand the answer...
I'm sorry to say that not only is 8C3 NOT obvious, it's also wrong.Originally posted by McLake
METHOD I:
8C3 is obvious
to include both e's selection is e,e,# and there are 6 hashes. Order is not important (yet).
so no e must be 50 (8C3 - 6)
so arrangments are 50 * 3! (cause can arrange three ways) + 6 x 3!/2! (casue can arrange three ways but 2 of them are same due to e's).
METHOD II:
we have 8 letters to choose from
then 7
then 6
so 8*7*6 = 336
but 6*3 are the same
so no of combo = 336 - 18 = 318