Answer
$P(10)=0.186$
Work Step by Step
$P(x)={}_nC_{x}~p^x~(1-p)^{n-x}$
n = 15
x = 10
p = 60% = 0.6
1 - p = 1 - 0.6 = 0.4
n - x = 15 - 10 = 5
$P(10)={}_{15}C_{10}\times0.6^{10}\times0.4^{5}=\frac{15!}{10!(15-10)!}\times0.6^{10}\times0.4^5=\frac{15!}{10!\times5!}\times0.6^{10}\times0.4^4$
But, $15!=15\times14\times13\times12\times11\times10\times9\times8\times7\times6\times5\times4\times3\times2\times1=15\times14\times13\times12\times11\times10!$
$P(10)=\frac{15\times14\times13\times12\times11\times10!}{10!\times5!}\times0.6^{10}\times0.4^5=\frac{15\times14\times13\times12\times11}{5\times4\times3\times2\times1}\times0.6^{10}\times0.4^5=0.186$
