Explanation:

Let’s break it down:

- We have 150 shooters in total, choosing at least one gun (A, B, C).

- 26 pick all three guns.

- 71 pick B (but not necessarily only B).

- 64 pick exactly two guns (this means: (A&B only) + (B&C only) + (A&C only) = 64).

- 48 pick both A and C (this is A n C, possibly overlapping with B).

- 27 pick only A.

- 50 pick A but not C (so, “only A” plus “A & B only”).

- For the bullets stuff—they’re just red herrings. Ignore them for the count of C.

So, let's find out how many people picked C.

Let’s use the set notation—with a bit of plain talk.

Let’s define:

- Only A = 27

- Only B = y

- Only C = z

- A & B only = p

- B & C only = q

- A & C only = r

- All three = 26

Given:

- A & C = r + 26 = 48 ? r = 22

- Exactly two guns = p + q + r = 64

- r = 22, so p + q = 42

Participants who pick A but not C: 50 = Only A + (A & B only) = 27 + p ? p = 23

Given p + q = 42 ? q = 19

So:

- Only A = 27

- Only B = y

- Only C = z

- A & B only = 23

- B & C only = 19

- A & C only = 22

- All three = 26

Sum all:

27 (A) + y (B) + z (C) + 23 + 19 + 22 + 26 = 150

So,

27 + y + z + 23 + 19 + 22 + 26 = 150

27 + 23 = 50

50 + 19 = 69

69 + 22 = 91

91 + 26 = 117

117 + y + z = 150 ? y + z = 33

Participants who pick B:

71 = Only B + (A & B only) + (B & C only) + (All three)

So,

71 = y + 23 + 19 + 26

23 + 19 + 26 = 68

71 - 68 = y = 3

So Only B = 3

Then z = 33 - 3 = 30

Now, people who pick C:

- Only C = 30

- B & C only = 19

- A & C only = 22

- All three = 26

So, total = 30 + 19 + 22 + 26 = 97

Here’s what this really means: the correct answer is Option 4, 97.

Summary of options:

- 95 (too low)

- 96 (close, but not right)

- 93 (wrong)

- 97 (Correct)

- 100 (too high)