As, nPr=720, then,
n!(n−r)!=720 (1)
And,
nCr=120
n!r!(n−r)!=120
From equation (1),
1r!×720=120
r!=6
r!=3×2×1
r!=3!
r=3
If nPr=720 and nCr=120, find the value of r.