For how many different ways, can you place 14 students in two rows of desks?

For how many different ways, can you place 14 students in two rows of desks, where 5 of them sit always in first row, and 4 of them sit always in second row? Eight students can sit in on row.

• Have you tried anything ? What is your specific difficulty ? – true blue anil Oct 3 '15 at 15:51
• I had only one idea. Split it into three parts: -in first row 5 children + 2 extra -in second row 4 childern + 3 extra -in first row 5 children + 1 extra -in second row 4 childern + 4 extra -in first row 5 children + 3 extra -in second row 4 childern + 2 extra so I got: $$\left ( \frac{8!}{3!} \cdot 3 \cdot \frac{8!}{4!}*4! \right ) + \left ( \frac{8!}{3!} \cdot 3 \cdot 2 \cdot \frac{8!}{4!}\cdot 4 \cdot 3 \cdot 2 \right ) + \left ( \frac{8!}{3!} \cdot 3 \cdot 2 \cdot \frac{8!}{4!} \cdot 4 \cdot 3 \right )$$ but it is totally wrong approach – EnemyPanda Oct 3 '15 at 19:41
• The numbers don't seem to add up, do they? You said there are 14 students, but you separate them into five who always sit in the first row, four who sit in the second row, and eight who sit in either(?) row. 4+5+8= ?? – hardmath Oct 4 '15 at 0:26
• What does "Eight students can sit in on row" mean ? Will you reproduce the quest ion exactly ? – true blue anil Oct 4 '15 at 2:48
• No, I see it like there are 2 desks, up to eight students can sit in every desk and at least 5 of them sit in first desk, and 4 of them sit in second desk – EnemyPanda Oct 4 '15 at 8:43

possible configurations are $(5+3)-(4+2),\; (5+2)-(4+3),\; (5+1)-(4+4)$
$\binom{5}{3}*^8P_8*^8P_6 + \binom{5}{2}*^8P_7*^8P_7 + \binom{5}{1}*^8P_6*^8P_8$
• In ${5\choose 3} * 8! * 6!$ Why do you permute in second row by 6? There are 6 students per 8 places – EnemyPanda Oct 4 '15 at 14:15