Question from Mac, a student:
Basically i like to know the difference between two solutions.
( I taken this questions from a web link)
Questions1) From 6 boys and 4 girls, 5 are to be selected to admission for a particular
house. In how many ways can this be done if there must be exactly 2 girls.
sol)
C(4,2)= 6
C(6,3)= 20
solutions is : 6*20= 120.
Question2) Out of 10 consonants and 4 vowels how many words can be formed each
containing 6 consonants and 3 vowels.
sol) C(10,6)
C(4,3)
and then 9! for arranging the selected words(9 character word).
solution= C(10,6) * C(4,3) * 9! = 30,48,19,200
Now my question is, why are we multiplying 9! in the second problem
and why we didn't multiply 5! in the first question ?


We have two responses for you
Hi Mac,
Up to a point the two problems are essentially the same.
Questions1) From 6 boys and 4 girls, 5 are to be selected to admission for a particular house. In how many ways can this be done if there must be exactly 2 girls.
sol)
C(4,2)= 6
C(6,3)= 20
Thus there are 6 × 20 = 120 ways to select the 5 people satisfying the conditions. 
Question2) Out of 10 consonants and 4 vowels how many words can be formed each containing 6 consonants and 3 vowels.
sol)
C(10,6) = 210
C(4,3) = 4
Thus there are 120 × 4 = 480 ways to select the 9 letters satisfying the conditions. 
But that's not the end of problem 2. Once you have the letters you need to form a word. There are many words that you can form from the 9 letters. You can choose the first letter in 9 ways and once you choose the first letter you can choose any of the 8 remaining for the second letter. Thus there are 9 × 8 ways to choose the first two letters. Continuing in this way there are 9 × 8 × 7 × ··· × 1 = 9! ways to form a word from the 9 letters you selected.
Hence the answer to problem 2 is 9! × 120 × 4.
Harley
In Question 2, I suppose you mean ``how many 9letter words can be formed each containing 6 DIFFERENT consonants and 3 DIFFERENT vowels''. Then C(10,6) * C(4,3) count the number of
SETS of 6 consonnants and 3 vowels to be used, and 9! counts the ways of making the word by then selecting a first letter, a second letter, a third letter.
To make such a selection needed in the first question, each of the five selected persons would have to have a special role, as in the following question:
Question3) From 6 boys and 4 girls, 5 are to be selected for a hockey team
with no goalie: One center, one leftwinger, one rightwinger, one right defender, one left defender. In how many ways can this be done if there must be exactly 2 girls.
sol)
C(4,2)= 6
C(6,3)= 20
solutions is : 6*20= 120.
