Question
Question: If the value of \(^{10}{{P}_{r}}=5040\), then the value of r is: [a] 2 [b] 3 [c] 4 [d] None ...
If the value of 10Pr=5040, then the value of r is:
[a] 2
[b] 3
[c] 4
[d] None of these
Solution
Hint: Start from r = 0 and check if LHS = RHS. If not put r =1 and check again. Repeat the process till you get LHS = RHS or r>10. If r>10 then there is no such value of r. Alternatively Divide RHS by 10, then 9 and continue till RHS = 1 or if RHS is not a natural number. If RHS is not a natural number, then there is no solution. Otherwise, the solution is 10 - the last number with which you divided +1.
Complete step-by-step answer:
We know that nPr=(n−r)!n!=n(n−1)…(n−r+1)
Now, put n = 10 and r =0 , we get
LHS =10P0=1
Since LHS = RHS, we continue
Put n = 10, r= 1
LHS =10P1=10
Since LHS = RHS, we continue
Put n = 10, r= 2
LHS =10P2=10×9=90
Since LHS = RHS, we continue
Put n = 10, r= 3
LHS =10P3=10×9×8=720
Since LHS = RHS, we continue
Put n = 10, r= 4
LHS =10P4=10×9×8×7=720=5040
Here LHS = RHS.
Hence r =4.
Hence option [c] is correct.
Note: Alternate solution:
RHS = 5040.
Dividing RHS by 10, we get RHS = 504
Since RHS =1, we continue
Dividing RHS by 9, we get RHS = 56
Since RHS =1, we continue
Dividing RHS by 8, we get RHS =7.
Since RHS =1, we continue
Dividing RHS by 7, we get RHS = 1.
Since RHS = 1, we stop.
Hence r = 10-7+1=4.
Hence option [c] is correct.