Pascal triangle just got better

What will be the value of the 1 0 th 10^\text{th} expression (applying operation from left to right).

First expression 1 Second expression 1 ÷ 1 Third expression 1 ÷ 2 ÷ 1 Fourth expression 1 ÷ 3 ÷ 3 ÷ 1 Fifth expression 1 ÷ 4 ÷ 6 ÷ 4 ÷ 1 Sixth expression 1 ÷ 5 ÷ 10 ÷ 10 ÷ 5 ÷ 1 Seventh expression 1 ÷ 6 ÷ 15 ÷ 20 ÷ 15 ÷ 6 ÷ 1 \begin{array} {lc} \text{First expression} & 1 \\ \text{Second expression} &1\div1 \\ \text{Third expression} & 1\div2\div1 \\ \text{Fourth expression} & 1\div3\div3\div1\\ \text{Fifth expression} & 1\div4\div6\div4\div1\\ \text{Sixth expression} & 1\div5\div10\div10\div5\div1\\ \text{Seventh expression} & 1\div6\div15\div20\div15\div6\div1 \end{array}

If you are getting answer like w . x y z × 1 0 a w.xyz\times10^{a} after correction to three decimal places, then enter it as w . x y z E + a w.xyzE+a

Clarification: E E represents E notation


The answer is 8.504E-14.

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1 solution

Alex G
Apr 26, 2016

The ( n + 1 ) t h (n+1)th row of Pascal's triangle is described as: ( n 0 ) \large{ {n} \choose {0}} ( n 1 ) \large{ {n}\choose {1}} ... ( n n 1 ) \large{ {n} \choose {n-1}} ( n n ) \large{ {n} \choose {n}} .

Using this, we get: ( 9 0 ) \large{ {9} \choose {0}} ÷ ÷ ( 9 1 ) \large{ {9} \choose {1}} ÷ . . . ÷ ÷...÷ ( 9 9 ) \large{ {9} \choose {9}} which can be evaluated to be 1 11759522374656 \frac {1} {11759522374656} or 8.503 E 14 8.503 E -14

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