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17 Jan 2022, 06:24
Kudos
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
70% (00:37) correct
30%
(00:46)
wrong
based on 77
sessions
History
Date
Time
Result
Not Attempted Yet
How many 2 digit numbers can be formed using any of the 10 digits when repetition is allowed?
(A) 50
(B) 70
(C) 90
(D) 100
(E) 180
(A) 50
(B) 70
(C) 90
(D) 100
(E) 180
17 Jan 2022, 06:35
Kudos
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chetan2u
Mugdho
I have rewritten the query you had asked here, as that post is locked.
Two ways as you too mentioned
1) Permutation:
Two places AB
A can be any of the digits except 0, so 9.
B can be any of the 10 digits
So 9*10 or 90
2) Combination
You have taken it as 9C1*10C1*2!. That is wrong.
Permutation will be choose any two digits in 10C2 or 45 ways and they can be arranged in 2! ways, so 45*2=90 ways where digits are different.
Add ways when both digits are same, so 10 ways.
But subtract ways where 0 is in tens place, so 1*10 or 10 ways
Total 90+10-10 = 90 ways.
17 Jan 2022, 07:32
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chetan2u
chetan2u
but why do we have to take 2 digits at a time(10c2)?
There are 10 digits.
So for ten's digit we can choose any one from the 9 digits and for unit digit we can choose any one digit from the 10 digits. And then we then arrange them (as we do for other problems of this kind). Why this idea is not valid?!
Posted from my mobile device
