[The Pigeonhole Principle (continued).]

In a photo studio, 20 birds flew in - 8 sparrows, 7 wagtails, and 5 woodpeckers. Each time the photographer clicks the camera shutter, one of the birds flies away (permanently). How many shots can the photographer take to be sure: he will have at least four birds of one species left, and at least three of another?

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8 snapshots are dangerous: in this time, 3 woodpeckers and 5 wagtails may fly away, leaving only 2 of each.

Let's show that 7 snapshots can be made. Then, in the studio, there will be $20-7=13$ birds left. This means that the number of birds of one species is at least $13: 3$, that is, at least 5. On the other hand, their number is no more than 8, so the number of the two remaining species together is at least 5, and, therefore, the number of one of these species is at least three.

## Answer

7 snapshots.