#### In how many ways $20$  tickets can be divided into $2$ groups such that all contain $10$ tickets to each.Option: 1 $92300$Option: 2 $18300$Option: 3 $18356$Option: 4 $92378$

The given information is:

Number of tickets $=20$

Number of teams $=2$

Number of tickets to each group $=10$

Pick first $10$ tickets from $20$ tickets,

Total no. of ways of choosing is:

$\frac{20 !}{(20-10) ! 10 !}=\frac{20 !}{10 ! 10 !}=184756$

Pick last $10$ tickets from $10$ tickets are and the number of ways this can be done is $1$ .

Since the two groups are similar, there is no differentiation between them. Therefore, we need to divide it with $2 !=2$ .

The total numbers of ways:

$\frac{184756}{2}=92378$
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 18, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}