Solution:

First, find the total number of ways to form the committee without the conflict restriction. We need to choose 2 men out of 6 and 2 women out of 5.
Total ways = $6C2$ $\times$ $5C2$ = $15 \times 10 = 150$.
Next, find the number of "unwanted" committees where both Mr. Smith and Ms. Jones are selected. If both are on the committee, we still need 1 more man (from the remaining 5) and 1 more woman (from the remaining 4).
Unwanted ways = $5C1$ $\times$ $4C1$ = $5 \times 4 = 20$.
Subtract the unwanted committees from the total possible committees: $150 - 20 = 130$.

So, the answer is B

Solution:

The word "SUCCESS" has 7 letters: 3 'S's, 2 'C's, 1 'U', and 1 'E'.
To ensure no two 'S's are adjacent, we first arrange the non-S letters (U, C, C, E).
Ways to arrange U, C, C, E = $\frac{4!}{2!} = 12$ ways.
When these 4 letters are arranged, they create 5 potential gaps (including the ends) where an 'S' can be placed: _ Letter _ Letter _ Letter _ Letter _
To place the 3 'S's into these 5 gaps, we use combinations: $5C3$ = $10$ ways.
Total valid arrangements = $12 \times 10 = 120$.

So, the answer is B

Solution:

First, choose the 2 letters that will go into their correct envelopes. This can be done in $5C2$ = $10$ ways.
For the remaining 3 letters, none can go into their correct envelopes. This is a "derangement" problem. The number of derangements for $n$ items is $D_n = n!(1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + ...)$.
For 3 items, $D_3 = 3!(\frac{1}{2!} - \frac{1}{3!}) = 6(\frac{1}{2} - \frac{1}{6}) = 3 - 1 = 2$ ways.
Total ways = (Ways to choose correct letters) $\times$ (Ways to derange the rest) = $10 \times 2 = 20$.

So, the answer is B

Solution:

First, select the 1 complete pair you want from the 6 available pairs. This can be done in $6C1$ = $6$ ways.
You still need to choose 2 more shoes, and they must NOT form a pair.
Choose 2 different pairs from the remaining 5 pairs to pull single shoes from: $5C2$ = $10$ ways.
For each of those 2 pairs selected, you must choose exactly one shoe (either the left or the right). That gives $2C1 \times 2C1 = 2 \times 2 = 4$ ways.
Total ways = $6 \times 10 \times 4 = 240$.

So, the answer is B

Solution:

We break the journey into two legs: $(0,0)$ to $(3,2)$, and then $(3,2)$ to $(6,4)$.
Leg 1: From $(0,0)$ to $(3,2)$ requires exactly 3 Right moves and 2 Up moves (5 total moves). The number of paths is $5C3$ (or $5C2$) = $10$.
Leg 2: From $(3,2)$ to $(6,4)$ requires moving $(6-3)=3$ Right and $(4-2)=2$ Up (5 total moves). The number of paths is again $5C3$ = $10$.
Total distinct paths = (Paths for Leg 1) $\times$ (Paths for Leg 2) = $10 \times 10 = 100$.

So, the answer is B

Solution:

To form a triangle, you need 3 non-collinear points.
The total number of ways to choose any 3 points from 12 is $12C3$ = $\frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220$.
However, selecting 3 points that lie on the same straight line will not form a triangle. The 5 collinear points can be chosen in $5C3$ = $10$ ways.
Subtract the invalid combinations from the total: $220 - 10 = 210$.

So, the answer is C

Solution:

First, seat the 4 women around the circular table. The number of ways to arrange $n$ distinct objects in a circle is $(n-1)!$.
Ways to seat the women = $(4-1)! = 3! = 6$ ways.
Once the women are seated, they create exactly 4 distinct gaps between them. The 4 men must sit in these 4 gaps to ensure the alternating pattern.
Because the seats are now defined relative to the specific women sitting next to them, this is treated as linear permutation: $4P4 = 4! = 24$ ways.
Total ways = $6 \times 24 = 144$.

So, the answer is A

Solution:

A number is divisible by 4 if its last two digits form a number divisible by 4.
From the digits {0, 1, 2, 3, 4, 5}, the possible 2-digit combinations divisible by 4 are: 04, 12, 20, 24, 32, 40, and 52. (7 cases total).
Case 1: The last two digits contain '0' (04, 20, 40). There are 3 such pairs.
For the remaining 2 front positions, we have 4 remaining digits with no restrictions. Ways = $4P2 = 12$. Total for Case 1 = $3 \times 12 = 36$.
Case 2: The last two digits do NOT contain '0' (12, 24, 32, 52). There are 4 such pairs.
For the remaining 2 front positions, the first digit cannot be '0'. Out of 4 remaining digits, only 3 are valid for the first spot. The second spot can then be any of the 3 remaining digits. Ways = $3 \times 3 = 9$. Total for Case 2 = $4 \times 9 = 36$.
Total valid numbers = $36 + 36 = 72$.

So, the answer is C

Solution:

We need to calculate the combinations of selecting people for each project in sequence.
First, choose 4 developers for Project A from the 10 available: $10C4 = 210$.
Next, choose 3 developers for Project B from the remaining 6: $6C3 = 20$.
Finally, assign the remaining 3 developers to Project C: $3C3 = 1$.
Because the projects are distinctly named (A, B, C), we do not divide by $2!$ to account for the two groups of 3 (unlike dividing identical groups).
Total ways = $210 \times 20 \times 1 = 4200$.

So, the answer is B

Solution:

To answer 7 questions while selecting at least 3 from each section, the student can only use the following combinations:
Scenario 1: 4 questions from Section X and 3 questions from Section Y.
Ways = $5C4 \times 5C3 = 5 \times 10 = 50$.
Scenario 2: 3 questions from Section X and 4 questions from Section Y.
Ways = $5C3 \times 5C4 = 10 \times 5 = 50$.
Since these scenarios are mutually exclusive, we add the possibilities together.
Total ways = $50 + 50 = 100$.

So, the answer is B