Four numbers are chosen at random (without replacement) from the set [ 1,2,3,....20 .] Sta…

Question

Four numbers are chosen at random (without replacement) from the set $$\left\{ {1,2,3,....20} \right\}.$$

Statement - 1: The probability that the chosen numbers when arranged in some order will form an AP is $${1 \over {85}}.$$
Statement - 2:  If the four chosen numbers form an AP, then the set of all possible values of common difference is $$\left( { \pm 1, \pm 2, \pm 3, \pm 4, \pm 5} \right).$$

Accepted Answer

Four numbers can be chosen ${}^{20}{C_4}$ ways.

When common difference  d = 1 then the possible sets are (1, 2, 3, 4) (2, 3, 4, 5) (3, 4, 5, 6) ................. (17, 18, 19, 20) = 17 sets

So when d = 1 then 17 different AP's are possible with 4 numbers.

Now let's create a table of all possible sets -

Common
Difference
(d)
    Possible Sets
    No of AP

d = 1
    (1, 2, 3, 4) (2, 3, 4, 5) (3, 4, 5, 6) ................. (17, 18, 19, 20)
    17

d = 2
    (1, 3, 5, 7) (2, 4, 6, 8) (3, 5, 7, 9) ................. (14, 16, 18, 20)
    14

d = 3
    (1, 4, 7, 10) (2, 5, 8, 11) (3, 6, 9, 12) ................. (11, 14, 17, 20)
    11

d = 4
    (1, 5, 9, 13) (2, 6, 10, 14) (3, 7, 11, 15) ................. (8, 12, 16, 20)
    8

d = 5
    (1, 6, 11, 16) (2, 7, 12, 17) (3, 8, 13, 18) (4, 9, 14, 19) (5, 10, 15, 20)
    5

d = 6
    (1, 7, 13, 19) (2, 8, 14, 20)
    2

$$\therefore$$ Total no of AP = 17 + 14 + 11 + 8 + 5 + 2 = 57

$$\therefore$$ Required probability = $${{57} \over {{}^{20}{C_4}}}$$ = $${1 \over {85}}$$

$$\therefore$$ Statement - 1: is true.

$$\therefore$$ Statement - 2: is false as common difference can also be $$ \pm 6$$.

Solutions · 1