Mathematics Test - 4

5 balls from out of 15 can be chosen in ¹⁵C₅ ways. The number of ways in which only one ball drawn will be white is ¹⁰C₁ × ⁵C₄. Therefore, the required probability is 

Since the exponent is 2n, there is only one middle term namely the [(2n/2) + 1] = (n + 1) th term i.e.

But we have T_{n + 1} = ²ⁿC_n xⁿ. So the coefficient of the middle term is:

There are 7! ways of arranging seven people in a row.

If A and B are the two people who initially set on the aisle, then after intermission A can sit in any of the

5 non-aisle seats and then B can sit in any of the remaining 4 non-aisle seats.

There are five other seats, and the remaining five people can sit in these set in 5! ways.

Thus, the probability is (5x4x5!)/7! = 10/21

(Ö3 + 1)⁴ + (Ö3 – 1)⁴ = (Ö3)⁴ + ⁴C₁ (Ö3)³ (1) + ⁴C₂ (Ö3)² + ⁴C₃ (Ö3) + ⁴C₄ + (Ö3)⁴ – ⁴C₁ (Ö3)³ (1) + ⁴C₂ (Ö3)² – ⁴C₃ Ö3 + ⁴C₄ 

= 2 [9 + 6(3) + 1] = 56 = a rational number.

Since the exponent is 2n, there is only one middle term namely the [(2n/2) + 1] = (n + 1) th term i.e.

But we have T_{n + 1} = ²ⁿC_n xⁿ. So the coefficient of the middle term is:

Since the last three digits are different, these can be chosen in 10 x 9 x 8 = 720 ways. Of these, exactly one choice will be the correct number. So, the required probability is 1/720 .