Each of 16 individuals is to be given an identifying code consisting o

hi,

TWO WAYS-

1) we can do by looking at choices too

3-digits

since 3 is the lowest lets see how many can be nad ewith 3..
1) single digit - 3
2) 2 digits - different digits - 3*2/2 = 3.... 12,13,23
3) 3-digits - 123 - only 1
total = 3+3+1 =7...

4-digits

1) single digit - 4
2) 2 digits - different digits - 4*3/2 = 6.... 12,13,14,23,24,34
3) 3-digits - 4*3*2/3! = 4
4) 4-digits - 4*3*2/4! = 1
Total = 4+6+4+1 = 15

so 5 should be the answer

C

2) Combinations formula...IMPORTANT
since in combination OREDER does not matter and when we place SOME digits in different order in Permutations, ONLY one out of them is in ascending order, we can work on Combinations
say total n digits are required
single digit will be nC1...
2 digits - nC2 and so on..
so we are looking for \(nC1+nC2+nC3+...nCn\geq{16}\)...
\(nC0+nC1+nC2+nC3+...nCn=2^n\) is a formula..
so \(nC1+nC2+nC3+...nCn=2^n-nC0=2^n-1\)..
so \(2^n-1\geq{16}.................2^n\geq{17}...................so.. n\geq5\)
so n=5

C

EDIT - MeghaP I have added the combinations solution too...