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

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(D) The number of times a prime $p$ divides $n!$ is given by Legendre's formula: $E_p(n!) = \sum_{k=1}^{\infty} \lfloor \frac{n}{p^k} \rfloor$. For $^

Vedclass · Accepted Answer

$29$

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(D) The number of times a prime $p$ divides $n!$ is given by Legendre's formula: $E_p(n!) = \sum_{k=1}^{\infty} \lfloor \frac{n}{p^k} \rfloor$. For $^{80}C_{40} = \frac{80!}{40!40!}$,the exponent of a prime $p$ is $E_p(80!) - 2E_p(40!)$. $(a)$ For $p=7$: $E_7(80!) = \lfloor \frac{80}{7} \rfloor + \lfloor \frac{80}{49} \rfloor = 11 + 1 = 12$. $E_7(40!) = \lfloor \frac{40}{7} \rfloor = 5$. Exponent in $^{80}C_{40} = 12 - 2(5) = 2$. Divisible by $7$. $(b)$ For $p=23$: $E_{23}(80!) = \lfloor \frac{80}{23} \rfloor = 3$. $E_{23}(40!) = \lfloor \frac{40}{23} \rfloor = 1$. Exponent in $^{80}C_{40} = 3 - 2(1) = 1$. Divisible by $23$. $(c)$ For $p=11$: $E_{11}(80!) = \lfloor \frac{80}{11} \rfloor = 7$. $E_{11}(40!) = \lfloor \frac{40}{11} \rfloor = 3$. Exponent in $^{80}C_{40} = 7 - 2(3) = 1$. Divisible by $11$. $(d)$ For $p=29$: $E_{29}(80!) = \lfloor \frac{80}{29} \rfloor = 2$. $E_{29}(40!) = \lfloor \frac{40}{29} \rfloor = 1$. Exponent in $^{80}C_{40} = 2 - 2(1) = 0$. Not divisible by $29$.

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

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