In how many different ways can 10 students be divided into 3 groups?


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Question Number 105551 by mr W last updated on 29/Jul/20

$I n h o w m a n y d i f f e r e n t w a y s c a n 10$ $s t u d e n t s b e d i v i d e d i n t o 3 g r o u p s ?$
	Answered by adhigenz last updated on 29/Jul/20

	$It is the same thing as finding how$ $many possible integer solution for$ $a + b + c = 10, which a, b, c ⩾ 1, since a$ $certain group must contain at least 1$ $student .$ $Let x = a - 1, y = b - 1, z = c - 1$ $We write the equation as :$ $x + y + z = 7, for x, y, z ⩾ 0$ $The solution of integer is (\begin{matrix} n + r - 1 \\ r - 1 \end{matrix})$ $= (\begin{matrix} 7 + 3 - 1 \\ 3 - 1 \end{matrix})$ $= (\begin{matrix} 9 \\ 2 \end{matrix})$ $= 36 ways$
	Commented by mr W last updated on 30/Jul/20

	$i {d o n}^{'} t t h i n k b o t h a r e t h e s a m e t h i n g .$ $10 s t u d e n t s a r e d i s t i n c t, s a y t h e y a r e$ $A, B, C, D, E, F, G, H, I, J .$ $s o f o r e x a m p l e$ $A, B, C, D / E, F, G / H, I, J$ $a n d$ $A, B, C, E / D, F, G / H, I, J$ $a r e t w o d i f f e r e n t w a y s .$
	Answered by 1549442205PVT last updated on 30/Jul/20

	$Consider a way of division into 3 groups$ $such that : one group of two, one group$ $of two, one group of six . Since for first$ $two persons we have C_{10}^{2} ways to choose,$ $for next two persons have C_{8}^{2} ways to$ $choose but because two - person repeated$ $. Hence all have \frac{C_{10}^{2} \times C_{8}^{2}}{2} = 45 \times 14 = 530$ $ways . Similarly, for way of division$ $(2, 3, 5) we have C_{10}^{2} \times C_{8}^{3} = 45 \times 56 = 2520 ways$ $iii) For (2, 4, 4) have \frac{C_{10}^{2} \times C_{8}^{4}}{2} = 45.35 = 1575$ $iv) For (3, 3, 4) have \frac{C_{10}^{3} \times C_{7}^{3}}{2} = 60.35 = 2100$ $Thus, total we have 530 + 2520 + 1575 + 2100$ $= 6725 ways to divide 10 student into$ $three groups .$
	Commented by mr W last updated on 30/Jul/20

	$b u t a g r o u p m a y a l s o h a v e o n l y o n e$ $s t u d e n t .$
	Commented by 1549442205PVT last updated on 30/Jul/20

	$If such as then adding$ $i) (1, 1, 8) have \frac{C_{10}^{1} \times C_{9}^{1}}{2} = 45 ways$ $ii) (1, 2, 7) have C_{10}^{1} \times C_{9}^{2} = 10.36 = 360 ways$ $iii) (1, 3, 6) have C_{10}^{1} \times C_{9}^{3} = 10.84 = 840 ways$ $iv) (1, 4, 5) have C_{10}^{1} \times C_{9}^{4} = 10.126 = 1260 ways$ $S = 45 + 360 + 840 + 1260 + 6725 = 9230$ $ways$
	Commented by mr W last updated on 30/Jul/20

	$t y p o i n r e s u l t f o r (2, 2, 6) : 45 \times 14 = 630$ $i n s t e a d o f 530!$ $t h e n y o u g e t S = 9330$ $t h i s i s c o r r e c t, n i c e w o r k s i r!$
	Commented by mr W last updated on 30/Jul/20

	$t h e c o e f f i c i e n t o f x^{10} t e r m f r o m$ $g e n e r a t i n g f u n c t i o n$ $\frac{10!}{3!} {(e^{x} - 1)}^{3} i s 9330 .$
	Commented by mr W last updated on 30/Jul/20

	Commented by 1549442205PVT last updated on 30/Jul/20

	$Thank you Sir! You are welcome .$
	Commented by mr W last updated on 31/Jul/20
	$number of ways to place n distinct objects in k identical boxes is {_k ^n } which is the stirling number of the second kind. {_k ^n }=S_2 (n,k) {_3 ^(10) }=9330.$
	$n u m b e r o f w a y s t o p l a c e n d i s t i n c t$ $o b j e c t s i n k i d e n t i c a l b o x e s i s {_{k}^{n}}$ $w h i c h i s t h e s t i r l i n g n u m b e r o f t h e$ $s e c o n d k i n d .$ ${_{k}^{n}} = S_{2} (n, k)$ ${_{3}^{10}} = 9330 .$
	Commented by mr W last updated on 31/Jul/20

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