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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 = $\dfrac2{77}$]

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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.

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  • $\begingroup$ I like your approach. In the second problem, we could just divide the six possible starting positions for the block by the $\binom{11}{6}$ ways we could choose the positions of the vowels. $\endgroup$ – N. F. Taussig Nov 23 '18 at 21:47
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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.

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  • $\begingroup$ Thankss, i did overlook e haha $\endgroup$ – Vanessa Nov 23 '18 at 21:42

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