Question

A bag contains 21 tickets bearing nos 1,2,3 .....21. If 3 tickets

are drawn at random, then the chance that they form A.P is

• A.
• B. $\frac { { } ^ { 10 } \mathrm { C } _ { 2 } + { } ^ { 11 } \mathrm { C } _ { 2 } } { { } ^ { 21 } \mathrm { C } _ { 3 } }$
• C. $\frac { { } ^ { 10 } \mathrm { C } _ { 2 } } { { } ^ { 21 } \mathrm { C } _ { 3 } }$
• D.

Answer 1

Answer: (B)

$\frac { { } ^ { 10 } \mathrm { C } _ { 2 } + { } ^ { 11 } \mathrm { C } _ { 2 } } { { } ^ { 21 } \mathrm { C } _ { 3 } }$

Answer 2

n(S) = ²¹C₃

let a, b, c be nos chosen

If a, b, c are in A.P, b = $\frac { a + c } { 2 }$

Ž a + c is even

a, c are both even

or a, c are both odd.

N(5) = no. of ways of selecting two evens or two odds

= ¹⁰C₂ + ¹¹C₂

P(5) = $\frac { { } ^ { 10 } \mathrm { C } _ { 2 } + { } ^ { 11 } \mathrm { C } _ { 2 } } { { } ^ { 21 } \mathrm { C } _ { 3 } }$ .