Method 1: We arrange the five letters B, K, K, P, R in a row, then insert the two O's and three E's in that order.
The number of ways we can arrange the five letters B, K, K, P, R in a row is
$$\frac{5!}{2!}$$
where we divide by $2!$ since we can permute the two K's within a given arrangement without producing an arrangement distinguishable from that arrangement.
We now have six spaces to fill, four between successive letters and two at the ends of the row. We can either place both O's in one of these six spaces, which can be done in $\binom{6}{1}$ ways, or place them in two different spaces, which can be done in $\binom{6}{2}$ ways.
Case 1: We place both O's in one of the six spaces.
This creates an arrangement with seven letters in which the two O's are adjacent. Since we are not permitted to have two adjacent O's, we must place an E in between the two O's. We now have an arrangement of eight letters, including the sequence OEO. This creates nine spaces, seven between successive letters and two at the ends of the row. Since the E's cannot be adjacent, we must separate them by placing an E in two of the seven spaces that are not adjacent to the E we have placed between the two O's. We can do this in $\binom{7}{2}$ ways. The number of such arrangements is
$$\binom{6}{1}\binom{7}{2}$$
Case 2: The two O's are placed in two different spaces.
This creates an arrangement of seven letters in which the O's are not adjacent. We have eight spaces to fill, six between successive letters and two at the ends of the row. To separate the E's, we choose three of these eight spaces in which to insert an E, which can be done in $\binom{8}{3}$ ways. The number of such arrangements is $$\binom{6}{2}\binom{8}{3}$$
Total: Since the two cases are disjoint, the number of arrangements of BOOKKEEPER in which the E's are not adjacent and the O's are not adjacent is
$$\frac{5!}{2!}\left[\binom{6}{1}\binom{7}{2} + \binom{6}{2}\binom{8}{3}\right] = 57960$$
Method 2: We use the Inclusion-Exclusion Principle.
There are $$\frac{10!}{2!3!}$$ distinguishable arrangements of the word BOOKKEEPER. From these we must exclude those arrangements in which two O's or two E's are adjacent.
One pair of adjacent identical letters:
Two O's are adjacent: We have nine objects to arrange: B, OO, K, K, E, E, E, P, R. Since the two K's are indistinguishable and the three E's are indistinguishable, they can be arranged in
$$\frac{9!}{2!3!}$$
distinguishable ways.
Two E's are adjacent: We have nine objects to arrange: B, O, O, K, K, EE, E, P, R. Since the two O's are indistinguishable and the two K's are indistinguishable, they can be arranged in
$$\frac{9!}{2!2!}$$
distinguishable ways.
Two pairs of adjacent identical letters:
Two O's are adjacent and two E's are adjacent: We have eight objects to arrange: B, OO, K, K, EE, E, P, R. Since the two K's are indistinguishable, they can be arranged in
$$\frac{8!}{2!}$$
distinguishable ways.
Two pairs of E's are adjacent: We have eight objects to arrange: B, O, O, K, K, EEE, P, R. Since the two K's are indistinguishable and the two O's are indistinguishable, they can be arranged in
$$\frac{8!}{2!2!}$$
distinguishable ways.
Three pairs of adjacent identical letters:
Two O's are adjacent and two pairs of E's are adjacent: We have seven objects to arrange: B, OO, K, K, EEE, P, R. Since the two K's are indistinguishable, they can be arranged in
$$\frac{7!}{2!}$$
distinguishable ways.
By the Inclusion-Exclusion Principle, the number of permissible arrangements is
$$\frac{10!}{2!2!3!} - \frac{9!}{2!3!} - \frac{9!}{2!2!} + \frac{8!}{2!} + \frac{8!}{2!2!} - \frac{7!}{2!} = 57960$$
Best Answer
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.
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.
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.
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.