combinatorics (1 Viewer)

[user18181818]

answer is c
why is it wrong to do 8! x 7 x 6 / 2! 2!
8! for middle terms (including grouped B), 7 for picking the first consonant and 6 for picking the last consonant. divide by 2! 2! since there are two B's and two I's?

 

[cossine]

answer is c
why is it wrong to do 8! x 7 x 6 / 2! 2!
8! for middle terms (including grouped B), 7 for picking the first consonant and 6 for picking the last consonant. divide by 2! 2! since there are two B's and two I's?
View attachment 40188

You need to pick the consonants first. All together there 6 consonants (technically 7 but we won't be able use B twice since it is joined together).

This gives 6*5. Note probability has eleven letters, 10 letters once we account for the fact regarding B. We have used two of the letters to deal with the consonant requirement. In other words we have 8 letters left. So we get 6*5*8!. However we need divide by 2! tp account for the double I.

This give use 6*5*8!/2 or in other words option C