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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. $54$
edited | 6.7k views
0
Can't we use permutation with repetition here?
–1

Will this logic be right ?

+1
For  1st position either 3 or 4 can be choosen in 2 ways

For 2nd position 2,3,4 can be choosen so 3  ways

For 3rd position 2,3,4 can be choosen so 3 ways

For 4th position 2,3,4 can be choosen so 3 ways

Total=2*3*3*3=54
+1
Notice -->

$\binom{9}{4}*4! - \binom{8}{3}*3!$ In this way you will get wrong answer because we are counting same number multiple times.

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 =$26 + 25 = 51.$

edited

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

0
in both the conditions its not clear why 3*3*3-2 and 3*3*3-1 coming.Please explain on this point kindly.

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
edited
0
Explain it
+1 vote

Explanation:

The given digits are 2, 2, 3, 3, 3, 4, 4, 4, 4 we have to find the numbers that are greater than 300

The first digit can be 3 or 4 but not 2.

Now, let us fix the first, second and third digits as 3, 2, 2, then the fourth place can be filled in 3 ways.

The number of ways is 3 similarly, we fix first third and fourth place as 3, 2 and 2 respectively (4) so the second place can be filled in 3 ways again,

The number of ways is 3

Now, we fix first, second and fourth, previous cases and we obtain the same result.

The number of ways is 3 so, the total number of ways is 9 similarly this can done by fixing the numbers as 3 and 4 (instead of 2) and thereby we obtain the a ways each

The number of numbers starting with 3 is 27

Similarly by taking 4 as the first digit we get 27 numbers

The number of numbers that are greater than 3000 is 27 + 27 = 54

But, 3222, 4222, is not possible as there are only two 2's, 3333 is not possible as there are only three 3's

The total number of numbers that are greater than 3000 is 54 - 3 = 51

Hope it helps u.:)

Numbers are 2,2,3,3,3,4,4,4,4

As it is 4 digit number first place must be 3,or 4

If 1st place is 3

2nd place can be fill up from any of rest 8 numbers.

3rd place can be fill up with any of rest 7 numbers

4 th place can be fill up with any of rest 6 numbers

So, total numbers $8\times 7\times 6=336$

If 1st place is 4

Similarly it also can be fill with 336 numbers

So, total $2\times 336$=672

+2
what if second and third places are filled with same digit? You are double counting them. Similarly for all positions.
0
Plz can any one explain why this method is going to be wrong

Simplest Approach Ever

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