20 students are numbered with the numbers from 1 to 20. 10 students are randomly selected. what is the probability that the sum of their numbers is exactly 100?


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Question Number 179699 by mr W last updated on 01/Nov/22

$20 s t u d e n t s a r e n u m b e r e d w i t h t h e$ $n u m b e r s f r o m 1 t o 20 . 10 s t u d e n t s a r e$ $r a n d o m l y s e l e c t e d . w h a t i s t h e$ $p r o b a b i l i t y t h a t t h e s u m o f t h e i r$ $n u m b e r s i s e x a c t l y 100 ?$
Commented by Rasheed.Sindhi last updated on 01/Nov/22

$◼ ∣ ◼ ∣ ◼ ∣ ◼ ∣ ◼ ∣ ◼ ∣ ◼ ∣ ◼ ∣ ◼ ∣ ◼ \cdot \cdot \cdot 100 ◼^{'} s & 99 ∣^{'} s$ $Now {we}^{'} ve to select 9 bars (divisions)$ $(or {we}^{'} ve to put 9 divisions (bars))$ $in a particular way which fulfills your$ $restrictions . . .$ $I remember this way or some type like$ $this which was introduced by sir$ $prakash jain / m r W s i r . . .$ $perhaps this method helps in such$ $questions . . . a l t h o u g h I^{'} m n o t s u r e .$
Commented by mr W last updated on 01/Nov/22

$t h e d i f f i c u l t y i s t h a t t h e s u m o f t h e$ $10 n u m b e r s s h o u l d b e 100 . w i t h ‘ ‘ s t a r s$ $a n d {b a r s}^{″} m e t h o d w e c a n n o t t r e a t t h e$ $s u m o f t h e n u m b e r s .$
Commented by JDamian last updated on 01/Nov/22

$p = \frac{c o m b i n a t i o n s s u m i n g u p 100}{a l l c o m b i n a t i o n s} ?$
Commented by mr W last updated on 01/Nov/22

$t h i s q u e s t i o n i s s i m i l a r t o Q 179420 .$ $b u t d u e t o t h e l a r g e n u m b e r 100, w e$ $c a n n o t e n u m e r a t e a l l c o m b i n a t i o n s$ $w h i c h g i v e t h e s u m 100 .$
Commented by Rasheed.Sindhi last updated on 01/Nov/22

$S i r i f w e p u t 9 b a r s b e t w e e n 100 s t a r s$ $t h e n w i l l t h e s u m o f t e n g r o u p s (m a d e$ $b y b a r s) n o t b e 100 ?$ $E x e c u s e m e i f I n o t g e t t o y o u .$
Commented by mr W last updated on 01/Nov/22

$t h i s i s t r u e . b u t h o w c a n y o u r e l a t e$ $t h e 100 s t a r s t o t h e n u m b e r s f r o m 1$ $t o 20 ?$ $a c t u a l l y w e j u s t w a n t t o d e t e r m i n e$ $h o w m a n y s o l u t i o n s f o l l o w i n g i n t e g e r$ $e q u a t i o n h a s :$ $n_{1} + n_{2} + . . . + n_{10} = 100$ $w i t h 1 ⩽ n_{1} < n_{2} < . . . < n_{10} ⩽ 20$
Commented by Rasheed.Sindhi last updated on 01/Nov/22

$Y e s s i r {y o u}^{'} r e r i g h t!$
Commented by mr W last updated on 01/Nov/22

$u s i n g g e n e r a t i n g f u n c t i o n w e f i n d o u t$ $5095 w a y s t o s e l e c t 10 n u m b e r s w h o s e$ $s u m i s 100 . s o t h e p r o b a b i l i t y i s$ $p = \frac{5095}{C_{10}^{20}} = \frac{5095}{184756} \approx 2 . 76 %$
Commented by mr W last updated on 01/Nov/22

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