Esame di Stato 2011 – 4

Il numero delle combinazioni di n oggetti a 4 a 4 è uguale al numero delle combinazioni degli stessi oggetti a 3 a 3.
Si trovi n.

Si tratta di risolvere $\displaystyle {n \choose 4} = {n \choose 3}$ , con $n \ge 4$ .

Quindi

$\displaystyle \frac{n!}{4!\ (n-4)!} = \frac{n!}{3!\ (n-3)!}$

$\displaystyle \frac{n!}{4!\ (n-4)!}\cdot \frac{(n-3)}{(n-3)} = \frac{n!}{3!\ (n-3)!}\cdot \frac {4}{4}$

$\displaystyle \frac{n!}{4!\ (n-3)!}\cdot (n-3) = \frac{n!}{4!\ (n-3)!}\cdot 4$

$\displaystyle \frac{n!}{4!\ (n-3)!}\cdot (n-3-4) = 0$

$\displaystyle \frac{n!}{4!\ (n-3)!}\cdot (n-7) = 0$

n = 7

Calcola i coefficienti binomiali

n	$\displaystyle {n \choose 4}$			$\displaystyle {n \choose 3}$
4	$\displaystyle {4 \choose 4}$	= $\displaystyle \frac{4!}{4!(4-4)!}$	= 1	$\displaystyle {4 \choose 3}$	= $\displaystyle \frac{4!}{3!(4-3)!}$	= 4
5	$\displaystyle {5 \choose 4}$	= $\displaystyle \frac{5!}{4!(5-4)!}$	= 5	$\displaystyle {5 \choose 3}$	= $\displaystyle \frac{5!}{3!(5-3)!}$	= 10
6	$\displaystyle {6 \choose 4}$	= $\displaystyle \frac{6!}{4!(6-4)!}$	= 15	$\displaystyle {6 \choose 3}$	= $\displaystyle \frac{6!}{3!(6-3)!}$	= 20
7	$\displaystyle {7 \choose 4}$	= $\displaystyle \frac{7!}{4!(7-4)!}$	= 35	$\displaystyle {7 \choose 3}$	= $\displaystyle \frac{7!}{3!(7-3)!}$	= 35

Genera le combinazioni e contale

n	$\displaystyle {n \choose 4}$	Combinazioni		$\displaystyle {n \choose 3}$	Combinazioni
4	$\displaystyle {4 \choose 4}$	`abcd`	1	$\displaystyle {4 \choose 3}$	`abc abd acd bcd`	4
5	$\displaystyle {5 \choose 4}$	`abcd abce abde acde bcde`	5	$\displaystyle {5 \choose 3}$	`abc abd abe acd ace ade bcd bce bde cde`	10
6	$\displaystyle {6 \choose 4}$	`abcd abce abcf abde abdf` `abef acde acdf acef adef bcde bcdf bcef bdef cdef`	15	$\displaystyle {6 \choose 3}$	`abc abd abe abf acd ace acf ade adf aef bcd bce bcf bde bdf bef cde cdf cef def`	20
7	$\displaystyle {7 \choose 4}$	`abcd abce abcf abcg abde` `abdf abdg abef abeg abfg acde acdf acdg acef aceg` `acfg adef adeg adfg aefg bcde bcdf bcdg bcef bceg` `bcfg bdef bdeg bdfg befg cdef cdeg cdfg cefg defg`	35	$\displaystyle {7 \choose 3}$	`abc abd abe abf abg acd ace acf acg ade adf adg aef aeg afg bcd bce bcf bcg bde bdf bdg bef beg bfg cde cdf cdg cef ceg cfg def deg dfg efg`	35