Question

Grace is training to be an airplane pilot and must complete five days of flying training in October with at least one day of rest between trainings. How many ways can Grace schedule her flying training in October (or any other 31-day month) given that she cannot train on consecutive days? ​Hint: consider the lengths of the gaps between flights.

Answer 1

Answer:

53130

Step-by-step explanation:

Let training days = T1, T2, T3, T4, T5

Let gaps in the days between two consecutive training days be =

                                                       

   G1      G2        G3        G4        G5      G6

   ↔T1,  ↔  T2,   ↔   T3, ↔  T4, ↔  T5 ↔            

Now   G1 & G6 ≥ 0   G1 ≥ 0, G6 ≥ 0

G2, G3, G4, G5 ≥ 1      G2 ≥ 1, G3 ≥ 1, G4 ≥ 1, G5 ≥ 1    

Therefore G2 -1 ≥ 0, G3 -1≥ 0, G4 -1 ≥ 0, G5 -1≥ 0

G1 + G2 + G3 + G4 + G5 + G6 = 25

G1 + (G2 -1) + (G3 -1) + (G4 -1) + (G5 -1) + G6 = 25-5

G1 + (G2 -1) + (G3 -1) + (G4 -1) + (G5 -1) + G6 = 20

Now the No. of non-negative integral = 20 + 6 – 1        = 25        =  53130

                                                                                          Ç6-1          Ç5

                                                                                               

                                                         

Therefore Grace can schedule her flying training in October in 53130 ways.