In how many arrangements of the letters of the word CALCULUS do both Ls appear before the first U?
Method 1: We choose two of the eight positions for the two Cs, one of the remaining six positions for the A, and one of the remaining five positions for the S. If every L must appear before the first U, the remaining four positions can be filled in only one way. Therefore, there are
$$\binom{8}{2}\binom{6}{1}\binom{5}{1} = 840$$
admissible arrangements.
Method 2: We use symmetry.
We choose two of the eight positions for the Cs, two of the remaining six positions for the Ls, two of the remaining four positions for the Us, and one of the remaining two positions for the A. The S must be placed in the remaining position. Hence, there are
$$\binom{8}{2}\binom{6}{2}\binom{4}{2}\binom{2}{1} = \frac{8!}{2!6!} \cdot \frac{6!}{2!4!} \cdot \frac{4}{2!2!} \cdot \frac{2!}{1!1!} = \frac{8!}{2!2!2!1!1!} = 5040$$
distinguishable arrangements of the letters of the word CALCULUS.
Within any given arrangement of the letters of the word CALCULUS, the two Ls and two Us can be permuted among themselves in $\binom{4}{2} = 6$ ways when the other four letters remain fixed since choosing two of the four positions occupied by the Ls and Us for the Ls also determines the positions of the Us. Only one of these six arrangements of the Ls and Us places both Ls before the first U. Thus, by symmetry, the number of admissible arrangements of the letters of the word CALCULUS is
$$\frac{1}{6} \cdot \frac{8!}{2!2!2!1!1!} = 840$$
which agrees with the answer obtained above and your unexplained answer.
In how many arrangements of the letters of the word CALCULUS does at least one L appear before the first U?
Method 1: We choose two of the eight positions for the two Cs, one of the remaining six positions for the A, and one of the remaining five positions for the S. The first of the remaining four positions must be filled with an L. That leaves three ways to place the remaining L. The remaining two positions must be filled with Us. Hence, the number of admissible arrangements is
$$\binom{8}{2}\binom{6}{1}\binom{5}{1}\binom{3}{1} = 2520$$
Method 2: We use symmetry.
We showed that there are $5040$ distinguishable arrangements of the word CALCULUS. Since there are two Ls and two Us in CALCULUS, by symmetry, in half of these arrangements, the first L appears before the first U. Hence, the number of admissible arrangements is
$$\frac{1}{2} \frac{8!}{2!2!2!1!1!} = 2520$$
which agrees with the answer obtained above.
In how many arrangements of the letters of the word CALCULUS do both Ls appear before the S?
Method 1: We choose two of the eight positions for the Cs, two of the remaining six positions for the Us, and one of the remaining four positions for the A. If every L must appear before the S, the remaining three positions can be filled in only one way. Therefore, there are
$$\binom{8}{2}\binom{6}{2}\binom{4}{1} = 1680$$
admissible arrangements.
Method 2: We use symmetry.
Within any given arrangement of the letters of the word CALCULUS, the two Ls and the S may be permuted among themselves in three ways when the other letters are fixed. Only one of these three arrangements of the Ls and S places both Ls before the S. Hence, by symmetry, the number of admissible arrangements is
$$\frac{1}{3} \cdot \frac{8!}{2!2!2!1!1!} = 1680$$
which agrees with the answer above and your unexplained answer.
In how many arrangements of the letters of the word CALCULUS does the first L appear before the S?
Method 1: We choose two of the eight positions for the Cs, two of the remaining six positions for the Us, and one of the remaining four positions for the A. Since the first L must appear before the S, there must be an L in the first open position. That leaves two ways to place the second L. The S must occupy the remaining open position. Hence, there are
$$\binom{8}{2}\binom{6}{2}\binom{4}{1}\binom{2}{1} = 3360$$
admissible arrangements.
Method 2: We use symmetry.
Since the word CALCULUS contains two Ls and one S, the first L appears before the S in $2/3$ of the arrangements. Hence, the number of admissible arrangements is
$$\frac{2}{3} \cdot \frac{8!}{2!2!2!1!1!} = 3360$$
which agrees with the answer obtained above.
Best Answer
Just following your logic. Say they are $X$ and $Y$ for argument sake -
Case(a): $XY$ together leads to $2 \times 4! = 48$ cases. If they are not distinct, they cannot be permuted within and so you need to subtract $24$ times.
Case (b):
$X \_ − − - \,$ has $3$ choices for $Y$ and so does $Y_ − − − -$ has $3$ choices for $X$ (total of $6$). If they are same, you overcounted $3 \times 3!$ times.
Similarly in $ \_ X \_ − − \,$ you overcount $2 \times 3!$ times.
$− \_ X \_ − $ has only one place of choice for $Y$ and similarly for $− \_ Y \_ − $, there is one choice for $X$. If they are same, you overcounted $1 \times 3!$ times.
If you add them up, you get $36$ cases you need to subtract from $(b)$.
So it is $ 120 - 36 - 24 = 60$.