Vrftsjtryk

Here's what I tried:

The total number of ways is $dfrac{11!}{3!cdot2!}$.

The consonants can be together in $dfrac{6(5)(9!)}{3!cdot 2!}$ ways. When I divide, I get $dfrac 3{11}$ but the answer is $dfrac2{11}$.... Where did I go wrong?

The second part states find the probability that all the vowels are together.

I did $7!times 10$ ($7$ if you consider all vowels as one unit, then multiply by $10$ cause you can rearrange vowels amongst themselves in $10$ ways), divided by the total number of ways, and I got $dfrac1{66}$... I don't see where I went wrong. [The answer = $dfrac2{77}$]

It's all about trying possible paths/combinations and their probabilities. For the first example, proceed sequentially:

1. Probability that the first letter that you pick is a consonant: $frac{5}{11}$ (5 consonants over eleven letters)


2. Probability that the second letter that you pick is a consonant: $frac{4}{10}$ (recall that you already picked one).

The probability you are looking for is $$frac{5}{11}*frac{4}{10}=frac{2}{11}$$

As for the second case, we can reason the same way. However, we will try a different argument. First, note that there are 11! possible combinations of the letters. Likewise, there are 6! ways to combine 6 elements and 5! possible ways to combine 5. Thus, the probability of getting any 6 particular letters together in a combination (for example, the six vowels) is $$6*,bigg(frac{6!,5!}{11!}bigg)=frac{1}{77}$$

The 6 at the beginning of the expression comes from the number of places (in an eleven-slot element) where you can have a sequence of 6 consecutive elements.

What is the probability that the first two letters are consonants if the letters of the word $EQUILIBRIUM$ are randomly arranged?

The eleven letters of the word $EQUILIBRIUM$ can be viewed as the multiset
$${1 cdot E, 1 cdot Q, 2 cdot U, 3 cdot I, 1 cdot L, 1 cdot B, 1 cdot R, 1 cdot M}$$
They include the six vowels $E, U, U, I, I, I$ and five consonants $Q, L, B, R, M$.

The number of distinct arrangements of the letters of the word $EQUILIRIUM$ is the number of ways we can select $3$ of the $11$ positions for the $I$s, two of the remaining $8$ positions for the $U$s, and then arrange the six letters that each appear once in the remaining six positions, which is
$$binom{11}{3}binom{8}{2}6! = frac{11!}{8!3!} cdot frac{8!}{6!2!} cdot 6! = frac{11!}{3!2!}$$
as you found.

If consonants occupy the first two positions, then the six vowels must be in the final nine positions. Choose three of those nine positions for the $I$s, two of the remaining six of those positions for the $U$s, and one of the remaining four of those positions for the $E$. Arrange the five consonants in the five open positions (including the first two positions from which we excluded the vowels). We can do this in
$$binom{9}{3}binom{6}{2}binom{4}{1}5!$$
ways.

Hence, the probability that the first two letters in a random arrangement of the letters of the word $EQUILIBRIUM$ are consonants is
$$frac{dbinom{9}{3}dbinom{6}{3}dbinom{4}{1}5!}{dbinom{11}{3}dbinom{8}{2}6!} = frac{2}{11}$$

Where did you make your mistake?

There are five consonants. Therefore, you should have obtained
$$5 cdot 4 cdot frac{9!}{3!2!}$$
favorable cases, which would have given you the probability
$$frac{5 cdot 4 cdot frac{9!}{3!2!}}{frac{11!}{3!2!}} = frac{2}{11}$$

What is the probability that all the vowels are together if the letters of the word $EQUILIBRIUM$ are randomly arranged?

If the six vowels are placed in a block, we have six objects to arrange: the block of vowels and the five consonants. The objects can be arranged in $6!$ ways. Within the block, we must choose three of the six positions for the $I$s and two of the remaining three positions for the $U$s. Hence, the number of arrangements in which all the vowels are together is
$$6!binom{6}{3}binom{3}{2}$$
Therefore, the probability that all six vowels are together if the letters of the word $EQUILIBRIUM$ are randomly arranged is
$$frac{6!dbinom{6}{3}dbinom{3}{2}}{dbinom{11}{3}dbinom{8}{2}6!} = frac{1}{77}$$
so there appears to be a typographical error in the answer key.

Where did you make a mistake?

It looks like you overlooked the fact that $E$ is a vowel. Had you accounted for that, you would have had
$$binom{6}{3}binom{3}{2} = 60$$
arrangements of the vowels within the block and $6!$ ways to arrange the block of vowels and five consonants.