First of all, C(14,4) is not 1001.
The only combinations possibilties are:
PPP (3 pencils)
PPF (2 pencils and 1 felt-tip)
PPB (2 pencils and 1 ballpoint, etc)
BBB
BBP
BBF
FFP
FFB (the least likely result)
PBF
Different permutations of the same group of three don't count as being different.
That is a total of only 9 possibilities.
The probabilities of each will not be the same.
A student has a cup with 14 writing implements: 7 pencils, 5 ball point pens, and 2 felt-tip pens.
In how many ways can the student select 4 writing implements? = C(14,4) = 1001
In how many ways can the selection be made if no more than one ball point pen is selected?
1 answer
1 answer