in how many ways can a teacher divide his 10 studens into 4 groups such that each group has at least 2 students?


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Question Number 212686 by mr W last updated on 21/Oct/24

$i n h o w m a n y w a y s c a n a t e a c h e r$ $d i v i d e h i s 10 s t u d e n s i n t o 4 g r o u p s$ $s u c h t h a t e a c h g r o u p h a s a t l e a s t 2$ $s t u d e n t s ?$
Commented by Spillover last updated on 21/Oct/24

$2268000 ?$
Commented by mr W last updated on 21/Oct/24

$h o w ?$
Commented by Spillover last updated on 21/Oct/24

$i s i t c o r r e c t ?$
Commented by Spillover last updated on 21/Oct/24

$\frac{10!}{{(2!)}^{4}} = 2268000$
	Answered by mehdee7396 last updated on 21/Oct/24

	$2 / 2 / 2 / 4 o r 2 / 2 / 3 / 3$ $\Rightarrow (\begin{matrix} 10 \\ 2, 2, 2, 4 \end{matrix}) + (\begin{matrix} 10 \\ 2, 2, 3, 3 \end{matrix}) = 44100$
	Commented by mr W last updated on 21/Oct/24

	$y o u r a n s w e r w e r e c o r r e c t i f t h e$ $q u e s t i o n i s t o d i v i d e t h e s t u d e n t s$ $i n t o 4 g r o u p s f o r m a t h m a t i c s,$ $p h y s i c s, c h e m i s t r y a n d b i o l o g y$ $r e s p e c t i v e l y, i . e . d i s t i n c t o b j e c t s$ $i n t o d i s t i n c t b i n s . b u t i n t h i s q u e s t i o n$ $w e j u s t d i v i d e 10 s t u d e n t s i n t o 4$ $(i d e n t i c a l) g r o u p s .$
	Commented by mehdee7396 last updated on 21/Oct/24

	$y o u a r e r i g h t \underset{―}{\underset{⏟}{⋚}}$
	Answered by mr W last updated on 21/Oct/24

	$2 / 2 / 2 / 4 \Rightarrow \frac{10!}{{(2!)}^{3} 3! 4!} = 3150$ $2 / 2 / 3 / 3 \Rightarrow \frac{10!}{{(2!)}^{2} 2! {(3!)}^{2} 2!} = 6300$ $Σ : 3150 + 6300 = 9450$
	Answered by Spillover last updated on 21/Oct/24

	$S t u d e n t = 10 (n) s a y$ $g r o u p = 4 (k) s a y$ $s e l e c t e d s t u d e n t i n e a c h g r o u p = (2 \times 4) = 8$ $10 - 8 = 2$ $N u m b e r o f a r r a n g e m e n t = (\begin{matrix} n + k - 1 \\ k - 1 \end{matrix})$ $(\begin{matrix} n + k - 1 \\ k - 1 \end{matrix}) = (\begin{matrix} 2 + 4 - 1 \\ 4 - 1 \end{matrix}) = (\begin{matrix} 5 \\ 3 \end{matrix}) = 10$ $T o t a l w a y = \frac{10!}{{(2!)}^{4}} \times 10 = 2268000$
	Commented by mr W last updated on 21/Oct/24
	$this is wrong sir. to divide 10 distinct objects into 4 identical bins such that no bin is empty, there are {_4 ^(10) }=S(10,4)=34105 ways. since in our question each bin should have at least 2 objects, so the answer should be less than 34105.$
	$t h i s i s w r o n g s i r .$ $t o d i v i d e 10 d i s t i n c t o b j e c t s i n t o 4$ $i d e n t i c a l b i n s s u c h t h a t n o b i n i s$ $e m p t y, t h e r e a r e {_{4}^{10}} = S (10, 4) = 34105$ $w a y s . s i n c e i n o u r q u e s t i o n e a c h b i n$ $s h o u l d h a v e a t l e a s t 2 o b j e c t s, s o t h e$ $a n s w e r s h o u l d b e l e s s t h a n 34105 .$
	Commented by mr W last updated on 22/Oct/24
	${_4 ^(10) } or S(10,4) is the stirling number of the second kind.$
	${_{4}^{10}} o r S (10, 4) i s t h e s t i r l i n g$ $n u m b e r o f t h e s e c o n d k i n d .$
	Answered by golsendro last updated on 22/Oct/24

	$(1) \frac{(\begin{matrix} 10 \\ 2 \end{matrix}) (\begin{matrix} 8 \\ 2 \end{matrix}) (\begin{matrix} 6 \\ 2 \end{matrix}) (\begin{matrix} 4 \\ 4 \end{matrix})}{3!}$ $(2) \frac{(\begin{matrix} 10 \\ 2 \end{matrix}) (\begin{matrix} 8 \\ 2 \end{matrix}) (\begin{matrix} 6 \\ 3 \end{matrix}) (\begin{matrix} 3 \\ 3 \end{matrix})}{{(2!)}^{2}}$
	Commented by golsendro last updated on 22/Oct/24

	$yes$
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