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...
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I don't understand this. Say we use 4 digits.
Then we can have 16 codes
1,2,3,4,12,13,14,21,23,24,31,32,34,41,42,43
Are these nos not distinct?
So it should be B.

Posted from my mobile device

out of 16 combinations you mentioned - 21 doesn't qualify the specified condition
that makes count to 15 combinations.
Adding a 5th digit will do.
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Why does 21 not qualify?
It is distinct from the rest and is placed in ascending order.
Kindly clarify

Posted from my mobile device

Hi,
Not only 21 but also 31, 41 do not qualify ..
as the digits are in ascending order..
12 is in ascending order but 21 is in descending order..
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Can you please explain a bit about this permutation:
2) 2 digits - different digits - 4*3/2
3) 3-digits - 4*3*2/3! = 4
might be trivial, but not getting it right now. How to get the combination for digits arranged in increasing order?

Hi,

its not trivial..
Many of us would not know that..

lets see 2 digits..
we have 4 digits - 1,2,3,4 - out of which we have to choose 2.....
half of these will be in increasing order and half not
order matters so 4*3... half of it 4*3/2..... 12 out of 12 and 21; 13 out of 13 and 31.. and so on..
4*3... MEANS 4 for the first place and remaining 3 for the 2nd digit

Now when we talk of 3 digits..
same rule applies we may make any ways but only ONE way will have all in ascending order..
so when we choose 123, we can make 123,231,132,213,312, and 321.. ONLY 123 is in ascending order...
so ONLY 1 out of 6 possible .. that is why we divide the TOTAL by 3! or 6...

ways of 3-digits out of 4 is 4*3*2....
But only 1 out of 6 is correct so 4*3*2/6...
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yes, Not only 21 but also 31, 41 do not qualify .. I overlooked it.
Thanks
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Is there a way to solve this using the combinations formula?

Hi MeghaP,
there is a combination way and have added to my solution above
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I worked on the numbers
1
12
123
23
13
1234
14
34
234
24
12345
15
25
25
45
345

We need at least 5 numbers.

C is the answer
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Thank you so much, really helpful!!

Here, instead of applying the formula of combinations, one could think and answer logically.

when we take 1, 2 and 3, it is not sufficient to create 16 codes- 1, 2, 3, 12, 13, 23, 123- thats it.

But when we take 1, 2, 3 and 4 we get the following codes:
1
2
3
4
12
13
14
23
24
34
123
124
134
234
1234

Well, I took two digits

so,

12,13,21,31,23,32 - that is using 2 digits that's max number of unique numbers. So 6 codes' using 3 numbers. Add one more
14,41,24,42,34,43 - So you get another 6 if you include another digit

All the numbers above can be ascending.

So to reach 16 you MUST have one more digit. Hence 5 digits. So answer is unique.

This is equivalent to selection as there is only one order and so nCr can be used.
Take 3 digits. 3C1+3C2+3C3=7 which is less than 16
Take 4 digits. 4C1+4C2+4C3+4C4=15 which is less than 16
So minimum number of distinct digits needed is 5.
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The code should have digits in ascending order.

In dire need of Kudos.

this question looks strange, but it is considered a gmat question by gmat4ready.
In math, this question is a word problem. The idea of this question is simple, but confusing.

OA is 5 b/c for {1,2,3,4,5}
We can have following codes: 1, 2,3,4,5, 12, 13, 14, 15, 123, 23, 24,...

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