GATE CSE 2010 | Question: 65

Question

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

• A. $50$
• B. $51$
• C. $52$
• D. $54$

Comments

Ans B. 51

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EASY APPROACH 
4 digit number :    _    _    _    _          & we want number > 3000

                           1st 2nd 3rd 4th

So, 1st place -> 2 ways (3 or 4)
2rd place -> 3 ways (2 or 3 or 4)

3rd place -> 3 ways (2 or 3 or 4)

3rd place -> 3 ways (2 or 3 or 4)
 

Total 4 digit numbers possible =  2 * 3 * 3 * 3 = 54
NOTE : 3 3 3 3  ,  3 2 2 2  , 4 2 2 2  [ NOT possible since #2's = 2 ; #3's = 3]
Desired 4 digit numbers = 54 - 3 = 51 

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when an artist is forced to study gate. Great explanation btw

Answer 1

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.

Comments

@amarVashishth Very nice explanation, thanks a lot.

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Helpful.

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there is an limitation of using 3 and 2 so while we are taking 3 we cannt take more than 3 so we will minus that case also while taking 3 or 4 we cannt take more than 2  2's so that will also be not taken into consideration

 

Answer 2

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

Comments

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

Answer 3

Simplest Approach Ever! 

Comments

can you explain more?

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nice approach.

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great👍

Answer 4

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

Comments

Explain it

Answer 5

Answer: Option B

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.:)

Answer 6

3 _ _ _

 

Filling with 2 3's

3 3 3 _

last place can be filled by a 2

this number can be altered in 3!/2! ways keeping the thousands place intact

last place can be filled by a 2

this number can be altered in 3!/2! ways keeping the thousands place intact

last place can be filled by a 4

this number can be altered in 3!/2! ways keeping the thousands place intact

so far 3+3

 

3 _ _ _

3 3 _ _

Two 2's in the gap →3!/2!

Two 4's in the gap→3!/2!

one two and one 4→3!

3+3+6=12

 

3_ _ _

3 2 _ _

Gaps can be filled by a two and a four →3!/2!

Gaps can be filled by 2 4's→3!/2!

3+3=6

 

3_ _ _

Filling all the gaps by 4→1 way only

 

3 3 3 2 3

3 3 3 4 3

3 3 2 2 3

3 3 4 4 3

3 3 2 4 6

3 2 2 4 3

3 2 4 4 3

3 4 4 4 1

so starting with 3 we have 6+12+6+1=25ways

 

4 _ _ _

4 4 4 4→1

 

4 3 4 4→3

4 3 2 4→6

4 3 2 2→3

4 3 3 2→3

4 3 3 4→3

4 3 3 3→1

 

4 2 4 4→3

4 2 3 4(counted)

4 2 3 3(counted)

4 2 2 4→3

4 2 2 3 (counted)

total 26 ways

 

51 ways total

 

Alternative approach:(Error prone)

 

Starting with 3 we have to fill 3 more spots

Consider each spot has 3 ways

3*3*3=27 ways but 3222 and 3333 not possible so subtract 2

25 ways

 

Starting with 4 we have 27 ways but 4222 is not possible so subtract 1

it's 26 ways