How many different 5-person subcommittees can be formed from a club having 10 members? (A)30,240 (B) 504 (C) 2,520 (d)252
can any ine help? is integrated math
??
how many ways are there to pick the first person?
or, if you have 10 pieces of candy in front of you; how many choices do you have on your first pick?
10?
yes :) and since youve taken 1 away, there are 9 left to choose from for the second pick ... 8 left for the 3rd pick .... you see a pattern in this yet?
yes i do
this selection is called "permuation", or in street talk its just "pick" 10 pick 5 is equal to 10*9*8*7*6 , the first 5 numbers from 10 but theres a catch to this value
so basically it will be 504 then.
there are groups from this pick that are "the same" that we need to divide out
it would take about 120 different permuations of 5 pick 5 to show the pattern which means that there are 120 duplications to divide out 30240/120 is what we are looking for
\[nCr=\frac{nPr}{r!}\]
so it willll turn out to be 252 if its divided
correct
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