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Find the number of rearrangements of the word CALCULUS such that the consonants are not all adjacent, no two vowels are adjacent, and the L's appear to the left of (but not necessarily adjacent to) the S.

S:Arrange the consonants in 5! 2! 3! = 10 ways, and then arrange the vowels in the 6 spaces between/before/after them in (6 * 5 * 4)/2! = 60 ways, for a total of 10 *60 = 600.

Question: then i know there are 5 consonants and 3 vowels but why 5!/2!3! ways and why 6 spaces and why 6*5*4/2 = 60?

Cooki Wang
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  • $\binom{5}{2}=\frac{5!}{2!3!}$ counts the number of ways of selecting two objects out of five. Here, this corresponds to the selection of which two of the five spaces occupied by consonants correspond to the letter C. The remaining three spaces will be occupied by the two L's and the S with the S specifically occupying the rightmost remaining available space. The $6$ refers to selecting which space between consonants corresponds to the letter A, and the $\binom{5}{2}=\frac{5\cdot 4}{2}$ corresponds to which two of the remaining spaces between consonants corresponds to the U's. – JMoravitz Nov 05 '17 at 00:49
  • i got what you said and one more thing is that for the 654/2 that the 2 represents.its 654 not 5*4 – Cooki Wang Nov 05 '17 at 01:42
  • The $\frac{6\cdot5\cdot4}{2}$ is explained by my last two sentences above. The two in particular comes in the part of the calculation of the binomial coefficient for the U's. – JMoravitz Nov 05 '17 at 01:44
  • yeah i see thank you but there is nothing in this answer have mentions anything that L's should be on the left side of S. right？ – Cooki Wang Nov 05 '17 at 01:46
  • That was accounted for in my third sentence in my first comment. There is nothing needing to be multiplied by here since there is exactly one way that the two L's and one S can be arranged within their three spaces such that the L's are to the left of the S. If you insist then you can multiply by one to help yourself sleep better at night, but there is no reason to visually include it in the final answer. – JMoravitz Nov 05 '17 at 01:48
  • im sorry i am still very confused, so can you explain to me by steps like how the solution solve the consonants are not all adjacents and how no 2 words are adjecent and how L's appear to the left of S. thank you – Cooki Wang Nov 05 '17 at 01:51

1 Answers1

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We first arrange the consonants by themselves in a line. We will later interweave the vowels inbetween the consonants.

$$\underline{~~~~}~~\underline{~~~~}~~\underline{~~~~}~~\underline{~~~~}~~\underline{~~~~}$$

We have five total consonants that we need to arrange: Two C's, two L's and one S.

We first pick which two of the five spaces to be used by the C's. This can be done in $\binom{5}{2}$ ways. For example like so:

$$\underline{~~~~}~~\underline{C}~~\underline{~~~~}~~\underline{~~~~}~~\underline{C}$$

In the remaining three spaces to be used by consonants, there will be two L's and one S. The problem statement tells us that the L's must occur before the S, so there is only one way to fill these spaces. Continuing our example, this would be:

$$\underline{L}~~\underline{C}~~\underline{L}~~\underline{S}~~\underline{C}$$

Now... we introduce the vowels. To do so, put a bit of extra space between each consonant for us to have the choice to put a vowel there. Once we are done placing the vowels, we will remove any leftover space.

$$\underline{~~~~}~~\underline{L}~~\underline{~~~~}~~\underline{C}~~\underline{~~~~}~~\underline{L}~~\underline{~~~~}~~\underline{S}~~\underline{~~~~}~~\underline{C}~~\underline{~~~~}~~$$

We choose which of the six available empty spaces will be occupied by the A. This can be done in six ways. For example like so:

$$\underline{~~~~}~~\underline{L}~~\underline{~~~~}~~\underline{C}~~\underline{~~~~}~~\underline{L}~~\underline{A}~~\underline{S}~~\underline{~~~~}~~\underline{C}~~\underline{~~~~}~~$$

We then choose which two of the remaining available spaces we place the U's. This can be done in $\binom{5}{2}$ ways. For example like so:

$$\underline{U}~~\underline{L}~~\underline{~~~~}~~\underline{C}~~\underline{~~~~}~~\underline{L}~~\underline{A}~~\underline{S}~~\underline{U}~~\underline{C}~~\underline{~~~~}~~$$

Removing the excess empty space we are left with the final arrangement, which in our example is $ULCLASUC$.

A few key points to notice here: When we decided to make space for the vowels and insert the vowels, we did so in a way that only one vowel could be in a space between two consonants at a time. This enforces our condition that no two vowels could be adjacent. Furthermore, as there are three vowels and at most one could be on the far left of the consonants and only one could be on the far right of the consonants, that implies that there must be a vowel in the middle of the consonants which automatically enforces the rule that the consonants cannot all be in one contiguous group for free with no extra effort needed on our part. That the Ls all occur before the S is taken care of in our process as well.

Applying multiplication principle, the total number of arrangements is then the product of the number of options at each step and is therefore

$$\binom{5}{2}\cdot6\cdot\binom{5}{2}=600$$

JMoravitz
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