In how many ways can we distribute 10 identical looking pencils to 4 students so that each student gets at least 1 pencil?

Question

ilovecake · Answer

210 is your answer 
yw

anonymous · Answer

$\Huge{\color{red}{\bigstar}\color{blue}{\bigstar}\color{green}{\bigstar}\color{yellow}{\bigstar}\color{orange}{\bigstar}\color{red}{\bigstar}\color{blue}{\bigstar}\color{green}{\bigstar}\color{yellow}{\bigstar}\color{orange}{\bigstar}\color{red}{\bigstar}\color{blue}{\bigstar}\color{green}{\bigstar}\color{yellow}{\bigstar}}\\color{white}{.}\\Huge\color{blue}{\mathfrak{~~~~Welcome~to~OpenStudy!~\ddot\smile}}\\color{white}{.}\\\Huge{\color{red}{\bigstar}\color{blue}{\bigstar}\color{green}{\bigstar}\color{yellow}{\bigstar}\color{orange}{\bigstar}\color{red}{\bigstar}\color{blue}{\bigstar}\color{green}{\bigstar}\color{yellow}{\bigstar}\color{orange}{\bigstar}\color{red}{\bigstar}\color{blue}{\bigstar}\color{green}{\bigstar}\color{yellow}{\bigstar}}$

paxpolaris · Answer

the 1st student can get get upto 7 pencils: 

if she/he gets 7: 1 for everyone else
$\large1$ way to do that. 

if she/he gets 6: 2pencils  for 1 student, 1 each for the other 2.
$\large3$ ways to do that. 

if 1st student gets 5: 3 for 1 student and 1 for other 2 (3+1+1). OR 2 for two students & 1 for one (2+2+1)
$3+3=\large6$ ways to do that. 

if 1st student gets 4: the others get (4+1+1) OR (3+2+1) OR (2+2+2)
$3+6+1=\large{10}$ ways to do that. 

if 1st student gets 3: the others get (5+1+1) OR (4+2+1) OR (3+2+2) OR (3+3+1)
$3+6+3+3=\large{15}$ ways to do that. 

if 1st student gets 2: the others get (7+1+1) OR (6+2+1) OR (4+2+2) OR (4+3+1) OR (3+3+2)
$3+6+3+6+3=\large{21}$ ways to do that. 

if 1st student gets 1: the others get (8+1+1) OR (7+2+1) OR (5+2+2) OR (5+3+1) OR (4+3+2) OR (4+4+1) OR (3+3+3)
$3+6+3+6+6+3+1=\large{28}$ ways to do that.

paxpolaris · Answer

Answer:
$$1+3+6+10+15+21+28=\Large \color{RED}{84}$$

paxpolaris · Answer

Another way of doing this: 

(7+1+1+1) :  $\large4$ ways to distribute like this
(6+2+1+1) :  $\large{12}$ ways
(5+3+1+1) :  $\large{12}$ ways
(5+2+2+1) :  $\large{12}$ ways
(4+4+1+1) :  $\large{6}$ ways
(4+3+2+1) :  $\large{24}$ ways
(4+2+2+2) :  $\large{4}$ ways
(3+3+2+2) :  $\large{6}$ ways
(3+3+3+1) :  $\large{4}$ ways
Total           : $\Large\overline{\color{red}{84}}$ ways