Find the number of different four digits number greater than 4000 such that the sum of their digits are odd

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2999

3000

2998

None of those

answer is B.

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Detailed Solution

It is given that the sum of the digits are odd.
⇒ Number can have a) 1 even digit, 3 odd digits
b) 1 odd digit, 3 even digits
Case I:- When number has 1 even digit, 3 odd digit
Subcase(a):-
When 1 even digit is at 1000th place:-
Since number are greater than 4000 and 1000th place should be even there are 3 possibilities i.e. 4,6,8
and in 100th, 10th, one place ‘now we can only fill with odd digits
i.e. say 100th place →can be filled with digits
1, 3, 5, 7, 9 → 5 ways
10th place → can be filled with digits
1, 3, 5, 7, 9 → 5 ways
Ones place → can be filled with digits
1, 3, 5, 7, 9 → 5 ways
Total number of ways in case (I) (a) are 3 × 5 × 5 × 5=375 ways
Sub-case (b):- When one even digit is at 100th, 10th or ones place i.e. in the middle
We know 1000th place can only be filled with odd digits which are greater than 4 since even digits is in the middle say 100th
So 1000th place → can be filled with digits is
5, 7, 9 → 3 ways
100th place → can be filled with even digits 2, 4, 6, 3, 0 → 5 ways only add.
Now 10th place → can be filled with digits
i.e. 1, 3, 5, 7, 9 → 5 ways only add and also ones place → can be filled with digits
1, 3, 5, 7, 9 → 5 ways
∴number of ways only even digit is at 1ooth place 3×5×5×5=375 ways but even can be at 10th and one place too
So total no. of ways = 375+375+375=1125ways.
Case II:- Subcase – (a) :- when 1 odd digit is at 1000th place .
Since the number are greater than 4000 and we have to fill odd digit at 1000th place
We have 3 case i.e.
1000th place can be filled with digits
5, 7, 9 -> 3 ways
and in the 100th, 10th& ones place we can now fill with only even digits.
i.e. say 100th place → can now be filled with digits.
2, 4, 6, 8, 0 → 5 ways
10th place →can now be filled with digits
2, 4, 6, 8, 0 → 5 ways
ones place → can now be filled with digits
2, 4, 6, 8, 0 → 5 ways
Total number of ways in case II subcases (a) are
= 3×5×5×5=375 ways

Subcase (b): when one odd digit is at 100th, 10th, or ones place i.e. in the middle.
We know now as one odd digit is in middle 1000th place can now be filled with only even digit which are greater than or equal to 4
1000th place can be filled with digits
4, 6, 8 => 3 ways
Now since one odd digit is in the middle say 100th
100th place → can be filled with digits
1, 3, 5, 7, 9 → 5 ways
Now 10th place → can now be filled with digits
2, 4, 6, 8, 0 → 5 ways
& one place → can now be filled with digits
2, 4, 6, 8, 0 → 5 ways
But this odd digit can also be at 10th& ones place.
So total number of ways in case –II subcases (b)
Are 3×3×5×5×5 = 1125.
∴∴total number of ways in which we can find different 4 digits number which are greater than 4000 & such that sum of digits are odd are
375+1125+375+1125
case I(a) case I(b) case II(a) case II(b)
= 3000 ways
Correct option is (2) 3000.

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