+471 Find the sum of all integers \(k\) such that \(\binom{23}{4} + \binom{23}{5} = \binom{24}{k}\).
+129852 C (23, 4) = 23! / [ (23 - 4)! * 4! ] = 23! / [ 19! * 4! ]
C(23,5) = 23! / [ (23 - 5)! * 5! ] = 23! / { 18! * 5!]
So.......we want to solve this
23! / [ 19! * 4! ] + 23! / [ 18! * 5!] = 24! / [ (24 - k)! * k! ]
1 / [ 19! * 24] + 1 / [ 18! * 120 ] = 24 / [ (24 - k)! * k! ]
[ 5 + 19 ] / [ 19! * 5!] = 24 / [ (24 - k)! * k! ]
Which implies that
19! * 5! = (24 - k)! * k!
Which implies that k = 5
But C (24, 5) = C(24, 19)
So...... 5 + 19 = 24
