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55 votes
55 votes

Given digits $ 2, 2, 3, 3, 3, 4, 4, 4, 4$ how many distinct $4$ digit numbers greater than $3000$ can be formed?

  1. $50$
  2. $51$
  3. $52$
  4. $54$
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9 Answers

Best answer
118 votes
118 votes

first place should be occupied by either $3$ or $4.$

Case 1 : First place is occupied by the digit $4$ 
 $4$  _  _  _
now in the set from where we can pick numbers is left with $=\{2,2,3,3,3,4,4,4\} $
if we got $3$ of each digit(which are $2, 3$ and $4$) then number of ways by each of those blanks can be

filled in are $3$ coz we have $3$ choices of digits: $\text{pick } 2, 3$ or $4.$
But we do not have just enough $2's$ to fill all those $3$ spaces with the digit $2.$

$\therefore$ we need to subtract this case where number would be $4222$.
So, total numbers obtained using the numbers in our current set $=1 \times 3 \times 3 \times 3 - 1 = 26$.

The first one is for the digit $4,$ coz its fixed for this case; the subtracted one is for the case $4222$ that can't be made possible.

Case 2: First place is occupied by the digit $3$ 
$3$  _  _  _
now in the set from where we can pick numbers is left with $=\{2,2,3,3,4,4,4,4\}$
we have enough $4's$ here but lack $3's$ and $2's$ $\therefore$, the cases we need to subtract are $3222$ and $3333$
So, total numbers obtained using the numbers in our current set $=1 \times 3 \times 3 \times 3 - 2 = 25$

both cases are independently capable of giving us the answer, we have =$ 26 + 25 = 51.$

Hence answer is option B.

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27 votes
27 votes

Answer is B 

Let first digit is 3 now rest 3 digits can be from 2,2,3,3,4,4,4,4 but in this no of 2's and 3's are only 2 so 222 and 333 is not possible so the possible no 3*3*3 -2 = 25

In second condition the First no is 4 now rest three digits can be 2,2,3,3,3,4,4,4 so here no of 2's is also two so possible numbers is 3*3*3* -1=26

25+26=51

5 votes
5 votes
answer - B

first digit can be 3 or 4

permutations for remaining 3 places as follows

no digit repeated = 3!

one digit repeated = 3!/2!

non repeating digit can be chosen in 2 ways = (3!/2!)2

repeating digit can be chosen in 3 ways = ((3!/2!)2)3

numbers greater than 3000 but less than 4000 = 3! + ((3!/2!)2)3  = 24 + 1(only 4 can be repeated 3 times

numbers grater than 4000 = 3! + ((3!/2!)2)3  + 2 (since 4 and 3 can be repeated 3 times) = 26

total numbers = 51
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