Question

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Answer 1

Din teorie stim ca:

$C^0_n+C^1_n+....+C^{n-1}_n+C^n_n=2^n$

In cazul de fata, n=5 si lipseste termenul C de 5 luate cate 0 (C de 5 luate cate 0=1) si avem:

$C^1_5+C^2_5+C^3_5+C^4_5+C^5_5=C^0_5+C^1_5+C^2_5+C^3_5+C^4_5+C^5_5-C^0_5=2^5-C^0_n=32-1=31$