^{80}C_{40} is not divisible by - | vedclass

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(D) The value of $^{80}C_{40}$ is given by $\frac{80!}{40! 	imes 40!}$.We use Legendre's Formula to find the exponent of a prime $p$ in $n!$,denoted

Vedclass · Accepted Answer

$29$

Vedclass · Answer

(D) The value of $^{80}C_{40}$ is given by $\frac{80!}{40! 	imes 40!}$.We use Legendre's Formula to find the exponent of a prime $p$ in $n!$,denoted by $E_p(n!) = \sum_{k=1}^{\infty} \lfloor \frac{n}{p^k} floor$.For $p=7$: $E_7(80!) = \lfloor \frac{80}{7} floor + \lfloor \frac{80}{49} floor = 11 + 1 = 12$. $E_7(40!) = \lfloor \frac{40}{7} floor = 5$. Thus,$E_7(^{80}C_{40}) = 12 - 2(5) = 2$. Divisible by $7$.For $p=23$: $E_{23}(80!) = \lfloor \frac{80}{23} floor = 3$. $E_{23}(40!) = \lfloor \frac{40}{23} floor = 1$. Thus,$E_{23}(^{80}C_{40}) = 3 - 2(1) = 1$. Divisible by $23$.For $p=11$: $E_{11}(80!) = \lfloor \frac{80}{11} floor = 7$. $E_{11}(40!) = \lfloor \frac{40}{11} floor = 3$. Thus,$E_{11}(^{80}C_{40}) = 7 - 2(3) = 1$. Divisible by $11$.For $p=29$: $E_{29}(80!) = \lfloor \frac{80}{29} floor = 2$. $E_{29}(40!) = \lfloor \frac{40}{29} floor = 1$. Thus,$E_{29}(^{80}C_{40}) = 2 - 2(1) = 0$. Not divisible by $29$.

$^{80}C_{40}$ is not divisible by -

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