Permutation Test - 6

The two elements P and Q such that P ∩Q can be chosen out of n is ⁿC₂ ways. Let a general element be a_i and must satisfy one of the following possibilities: 
(i) a_i∈ P and a_i∈Q   (ii) a_i∈ P and a_i∉Q 
(iii) a_i∉ P and a_i∈Q   (iv) a ­_i∉ P and a_i∉Q 
Let a₁, a₂∈ P ∩ Q 
only choice (i) a_i∈ P and a_i∈Q satisfy a₁, a₂∈P ∩ Q and three choices (ii), (iii) and (iv) for each of remaining (n – 2) elements. 
∴ No of ways of remaining elements = 3^{(n – 2)} 
Hence, No. of way sin which P ∩ Q contains exactly two elements = ⁿC₂ × 3^{(n – 2)}

Given, 10 Buttons, 10 Bulbs 
Required no. of ways to Enlighten all bulbs using either one button or Two buttons or three buttons or …… or ten buttons. 
= ¹⁰C₁ + ¹⁰C₂ + ¹⁰C₃ + ¹⁰C₄ + … + ¹⁰C₁₀ 
= ¹⁰C₀ + ¹⁰C₁ + ¹⁰C₂ + ¹⁰C₂ + ¹⁰C₃ + … + ¹⁰C₁₀– ¹⁰C₀ 
= 2¹⁰ – 1 
= 1024 – 1 
= 1023 
∴ Option B is correct answer.

Total number of spaces between n-3 people arranged in a row = n-2.

Therefore, number of ways of selecting 3 people out of n such that no 2 of them are consecutive = P_n = ^n–2C₃

The only difference in the second case is that the ends are not defined. Now, out of n-4 spaces, the number of ways of choosing one of them as an 'end' is ^n–4C₁

Q_n =^n–2C₃ - ^n–4C₁ 

Given that P_n – Q_n = 6 
⇒^n–4C₁ = 6

⇒ n = 10