If 5 different things are placed at random in 3 different boxes, then the probability of placing them such that no box remains empty is

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Solution

The correct option is C

n(s)= $3^{5}$ = 243

n(E) = $3 (^{5} C_{2} .^{3} C_{2} .^{3} C_{1})$ + $3 (^{5} C_{1} .^{4} C_{1} .^{3} C_{3}) = 150$

P(E)= $\frac{n (E)}{n (S)}$ = $\frac{50}{81}$

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If 5 different things are placed at random in 3 different boxes, then the probability of placing them such that no box remains empty is

The correct option is C n(s)=35= 243 n(E) = 3(5C2.3C2.3C1) + 3(5C1.4C1.3C3)=150 P(E)=n(E)n(S)=5081

The correct option is C

n(s)= $3^{5}$ = 243

n(E) = $3 (^{5} C_{2} .^{3} C_{2} .^{3} C_{1})$ + $3 (^{5} C_{1} .^{4} C_{1} .^{3} C_{3}) = 150$

P(E)= $\frac{n (E)}{n (S)}$ = $\frac{50}{81}$