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Taking a short cut through Pascal's triangle

机译:捷径穿过Pascal的三角形

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Everybody knows the proper way to travel through Pascal's triangle, starting at the top and moving down either left (L) or right (R), with each number in the triangle giving the total number of ways of reaching that position. But why restrict ourselves to left and right moves only, zigzagging our way down the triangle, when we could be going much more directly? We are crying out for vertical moves downwards to the next number below (that is, two rows lower). Let us call this sort of move D and consider the effect of having L, R and D on the number of possible ways of reaching each element. It somehow seems fitting that in a triangle there should be three possible moves. Putting a 1 at the top and continuing down gives us what I will call the short-cut triangle, the first few rows of which are shown Figure 1.
机译:每个人都知道通过Pascal三角形的正确方法,该方法从顶部开始,然后向左(L)或向右(R)向下移动,三角形中的每个数字给出到达该位置的总数。但是为什么当我们可以更直接地前进时,为什么只限制自己向左和向右移动,沿着三角形向下弯曲呢?我们渴望垂直向下移动到下面的下一个数字(即低两行)。让我们称这种动作为D,并考虑让L,R和D对到达每个元素的可能方式的数量的影响。以某种方式似乎合适的是,在三角形中应该有三个可能的移动。在顶部放一个1,然后继续向下,我们便称为快捷三角形,其前几行如图1所示。

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