Let there be a cube with n sides denoted 1 , . . . , n each. The cube is tossed n + 1 times. For 1 ≤ k ≤ n what is the probability that exactly k first tosses give different number (i.e, the ( k + 1 )-st toss give a number that was already gotten.) I really need to know why I got a slightly different answer from the official one.My attempt: Let us build a uniform sample space. Ω = { a i = ( i 1 , . . . , i k ) | 1 ≤ i j ≤ n }. | Ω | = ( n + 1 ) n , ∀ ω ∈ Ω , P ( ω ) = 1 | Ω | . We seek for the event A = { ( i 1 , . . . , i k , i k + 1 , . . . , i n + 1 ) | i t ≠ i s , ∀ 1 ≤ t ≠ s ≤ k , k ∈ { i 1 , . . . , i k } } This is the problematic part: | A | = ( n k ) ⋅ k ! ⋅ k ⋅ n n − k − 1 . (Then I and the answer use the formula for probability of an even it a uniform sample space.)The point is, the answer says: | A | = ( n k ) ⋅ k ! ⋅ k ⋅ n n − k I don't understand why; First I pick k numbers, count all their permutations, then pick one of them for the ( k + 1 )-th toss, and then I have n − k − 1 tosses left, each of which has n possibilities.

Question

Let there be a cube with   n sides denoted   1  ,  .  .  .  ,  n  each. The cube is tossed   n  +  1 times. For   1  ≤  k  ≤  n  what is the probability that exactly   k first tosses give different number (i.e, the   (  k  +  1  )-st toss give a number that was already gotten.) I really need to know why I got a slightly different answer from the official one.My attempt: Let us build a uniform sample space.   Ω  =  {      a    i    =  (      i    1    ,  .  .  .  ,      i    k    )  |  1  ≤      i    j    ≤  n  }.   |  Ω  |  =  (  n  +  1      )    n  ,   ∀  ω  ∈  Ω  ,  P  (  ω  )  =      1              |      Ω          |      . We seek for the event   A  =  {  (      i    1    ,  .  .  .  ,      i    k    ,      i          k      +      1        ,  .  .  .  ,      i          n      +      1        )  |      i    t    ≠      i    s    ,  ∀  1  ≤  t  ≠  s  ≤  k  ,  k  ∈  {      i    1    ,  .  .  .  ,      i    k    }  } This is the problematic part:   |  A  |  =                (              n      k                  )        ⋅  k  !  ⋅  k  ⋅      n          n      −      k      −      1      . (Then I and the answer use the formula for probability of an even it a uniform sample space.)The point is, the answer says:   |  A  |  =                (              n      k                  )        ⋅  k  !  ⋅  k  ⋅      n          n      −      k       I don&#039;t understand why; First I pick   k numbers, count all their permutations, then pick one of them for the   (  k  +  1  )-th toss, and then I have   n  −  k  −  1 tosses left, each of which has   n possibilities.

Sage Mcdowell · Accepted Answer

After toss k+1, you have n-k tosses remaining. (Also,   Ω  =  {  (      i    1    ,  …  ,      i          n      +      1        )  |  1  ≤      i    j    ≤  n  } so   |  Ω  |  =      n          n      +      1      ).The discrepancy is resolved by noting that the cube is cossed n+1 times, not n times.

Let there be a cube with n sides denoted 1 , . . . , n each. The cube is tossed

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