This is a counting question. Look at both diagrams, and see how many birds can be in.

If W is in, then W, G, and S are in. That’s 3 birds.

On this diagram, S, J, M and H are can be in. That’s 4. **C **is **CORRECT.**

Yes, you read that right. S can be in. Here’s how.

Start with J *in*. Don’t start with S *out* – you’re trying to put birds in!

If J is in, H is in. You can also put M in. J, M and H make three.

Only S is left. You don’t know *anything *about S if you start with J. Remember, you can only read the diagram left to right.

You can also put S in. Both S and J can be in together. S, J, M and H make four.

Review the setup if you’re not sure why S and J can both be in.

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Ravi Sandhu says

Why can’t all six be in? If S were in, that would fail the sufficient and make everything downstream technically irrelevant. It wouldn’t be triggered. I know that I am wrong; however, I would like to understand when a sufficient or necessary failing means that other relationships do not matter versus when other relationships continue to apply. Thank you so much.

TutorRosalie (LSATHacks) says

You can refer back to the setup page for a more comprehensive explanation. If S were in, it would not make everything downstream irrelevant because of the way it is related to J. The relationship says that if J isn’t in, then S is. This means that S is required for J to not be in. So it isnt’ the case that they can’t be in together, but that for J to be allowed to be out, S has to be there.

So we can have 3 cases: J and S both in, J in and S out, and S in and J out. So in this case, S doesn’t trigger anything, but elements down the rest of the chain would trigger for some to be out. For example, if J is in, then Rule 2 states that H must be in. Rule 1 and Rule 3 together then say that G and and W would then be out.