If four 2s make eight, and three 2s also make eight, how many 2s do you need to make three eights, when you only have less than eight 2s?
I take three eights as twenty-four, and can do $(2+2)(2+2+2)$ to get there with five twos. One or two of the plus signs can become multiplies. There are other solutions, depending on what operations are allowed. If we add division, we can multiply the above by $\frac 22$ and have solutions with seven twos. If concatenation and factorials are allowed, there are $22+2$ with only three, and $(2+2)!$ with just two.