Latest MBA Aptitude Question SOLUTION: How many 6 digit numbers contain exactly 4 different digits ?Plz post your approach..

1) 294840
2) 342422
3) 123421
4) 223344

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(#M40016604) MBA question permutations and combinations, probability Keep an EYE

How many 6 digit numbers contain exactly 4 different digits ?Plz post your approach..

1) 294840
2) 342422
3) 123421
4) 223344

Asked In MBA (11 years ago)

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AVIJIT  (10 years ago)

Let the four different digits be a, b, c, d, then the number can be of two forms i.e. aabbcd or aaabcd.

So to calculate the total numbers; first we need to find the number of ways to select 4 digits (which is C(10, 4)); then number of ways to select the digits to be repeated in each case (which is C(4,2) and C(4,1) respectively); and then the number of ways of arrangement in both the cases (which is 6!/(2!)2 and 6!/3! respectively).

Hence the number of ways to form such 6-digit numbers become = C(10, 4)[C(4, 2)6!/(2!)2 + C(4, 1)6!/3!] = 10*9*8*7*5*13 = 327600 .

But is this the final answer? .....certainly not

See in the initial selection for the digits, we have used all 10 digits (including 0) for selecting the four distinct digits. That means in the above arrangements there will be some numbers which'd be starting with zero. Now thinking alternately, first place of the number can be filled with any 10 digits equally. So removing the cases starting with zero, we are left with (9/10) of previously calculated result.

Thus the answer will be 9*9*8*7*5*13 = 294840.

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Dina Haines  (7 years ago)

To form a six digit number using exactly 4 different digits, we have to consider two cases:

1) Two digits are repeated and the other two are not (eg: 734573). This is possible in = 10C4 x 4C2 x 6!/2!2!

2) One digit is repeated thrice and the other three are not (eg: 734577). This is possible in = 10C4 x 4C1 x 6!/3!

Total ways = 10C4 x 4C2 x 6!/2!2! + 10C4 x 4C1 x 6!/3! = 226800+100800 = 327600

Since , 10% of these numbers would start with a zero.

So, answer is = 0.9 x 327600 = 294840

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