Trombone Trouble

$Let\quad there\quad be\quad 5\quad places\quad as\quad shown\quad below:\\ \quad \quad \quad \_ \_ \_ \_ \quad \_ \_ \_ \_ \quad \_ \_ \_ \_ \quad \_ \_ \_ \_ \quad \_ \_ \_ \_ \\ Since\quad the\quad basses\quad are\quad fixed\quad at\quad the\quad far\quad ends$

$SO\quad the\quad way\quad to\quad arrange\quad the\quad tenors\quad in\quad the\quad middle\\ 3\quad places\quad can\quad be\quad given\quad by\quad P(3,3)\quad =\quad 3!\quad =\quad 6\\ \\ Also\quad 2\quad basses\quad can\quad be\quad arranged\quad at\quad the\quad extreme\\ positions\quad in\quad 2\quad ways.\\ So,\quad by\quad fundamental\quad theorem\quad of\quad Arithmetic,\\ Total\quad number\quad of\quad ways\quad is\quad equal\quad to\quad 6\times 2=12$

There are 5 spots. _ x_x _x _x _

Basses must be on the end so.. 2 x _x _x _x 1

2 choices for first spot, leftover bass on the end.

Then three possible spots for 3 trombonists:

2 x 3 x 2 x 1 x 1 =12

Number of ways you can arrange the basses: 2! = 2

• (They can basically swap over on the two outermost spaces)

Number of ways you can arrange the tenors: 3! = 6

• (They are only able to fill the middle three spaces)

Therefore, total number of ways you can arrange the players: 2 x 6 = 12 arrangements