PonieważIle dzielników naturalnych ma \({27\choose11}\)?
- \(27!=2^{23}\cdot3^{13}\cdot5^6\cdot7^3\cdot11^2\cdot13^2\cdot17\cdot19\cdot23\),
- \(11!=2^8\cdot3^4\cdot5^2\cdot7\cdot11\),
- \(16!=2^{15}\cdot3^6\cdot5^3\cdot7^2\cdot11\cdot13\),
\[m={27\choose11}=\frac{2^{23}\cdot3^{13}\cdot5^6\cdot7^3\cdot11^2\cdot13^2\cdot17\cdot19\cdot23}{(2^8\cdot3^4\cdot5^2\cdot7\cdot11)\cdot(2^{15}\cdot3^6\cdot5^3\cdot7^2\cdot11\cdot13)}=3^3\cdot5^1\cdot13^1\cdot17^1\cdot19^1\cdot23^1\]
i
\[\overline{\overline{D_m}}=(3+1)\cdot(1+1)\cdot(1+1)\cdot(1+1)\cdot(1+1)\cdot(1+1)=2^7=128.\]
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