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Consider the following statements:
I. The number of positive integral solutions of $x_1+x_2+x_3+x_4=10$ is 286 .
II. If $25 !=10^n \times k,(k \in \mathbf{N})$, then $n=6$
Which one of the following options is true?
Options:
I. The number of positive integral solutions of $x_1+x_2+x_3+x_4=10$ is 286 .
II. If $25 !=10^n \times k,(k \in \mathbf{N})$, then $n=6$
Which one of the following options is true?
Solution:
2556 Upvotes
Verified Answer
The correct answer is:
Only II is true
The number of positive integral solution of
$x_1+x_2+x_3+x_4=10 \text { is }{ }^9 C_3=\frac{9 \times 8 \times 7}{1 \times 2 \times 3}=84$
$\therefore$ Statement I is false.
The exponent of 2 in 25 ! is
$\left[\frac{25}{2}\right]+\left[\frac{25}{4}\right]+\left[\frac{25}{8}\right]+\left[\frac{25}{16}\right]=12+6+3+1=22$
The exponent of 5 in (25)! is
$\left[\frac{25}{5}\right]+\left[\frac{25}{25}\right]=5+1=6$
$\therefore$ Exponent of 10 in $25 !=6$
Hence, $(25) !=10^6 \times k \Rightarrow k \in \mathbf{N}$
$\therefore$ Statement II is true.
$x_1+x_2+x_3+x_4=10 \text { is }{ }^9 C_3=\frac{9 \times 8 \times 7}{1 \times 2 \times 3}=84$
$\therefore$ Statement I is false.
The exponent of 2 in 25 ! is
$\left[\frac{25}{2}\right]+\left[\frac{25}{4}\right]+\left[\frac{25}{8}\right]+\left[\frac{25}{16}\right]=12+6+3+1=22$
The exponent of 5 in (25)! is
$\left[\frac{25}{5}\right]+\left[\frac{25}{25}\right]=5+1=6$
$\therefore$ Exponent of 10 in $25 !=6$
Hence, $(25) !=10^6 \times k \Rightarrow k \in \mathbf{N}$
$\therefore$ Statement II is true.
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