21 lines
628 B
Markdown
21 lines
628 B
Markdown
#Math #Probability
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# Problem
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Why does n choose k, or $\frac{n!}{k!(n-k)!}$ generate the coefficient for $x^ky^{n-k}$ in $(x+y)^n$?
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# Explanation
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Let’s see what happens when expanding $(x+y)^4$:
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$$
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(x+y)^4\\
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=(x+y)(x+y)(x+y)(x+y)\\
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=xxxx+\\
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yxxx+xyxx+xxyx+xxxy+\\
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yyxx+yxyx+yxxy+xyyx+xyxy+xxyy+\\
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xyyy+yxyy+yyxy+yyyx+\\
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yyyy
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$$
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When expanding, notice the number of terms with k of x (and likewise 4-k of y) is the number of combinations of 4 choose k, as you choose k slots to put k x’s in out of 4 slots. Therefore, $(x+y)^n={n \choose 0}x^0y^n+{n \choose 1}x^1y^{n-1}...+{n \choose n-1}x^{n-1}y^1+{n \choose n}x^ny^0$ |