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

SIMPLEST METHOD (This is how you should solve in GATE )


Assume an unlimited supply of 2, 3, and 4.

For the number to be greater than 3000, the first digit must be either 3 or 4.

• First digit = 3:

  3 _ _ _

  Remaining three positions each have 3 choices ⇒  3^3 =  27

• First digit = 4:

  4 _ _ _

  Again, 3^3 =  27

Thus,

Total = 27 + 27 = 54

But there are not unlimited 2,3 and 4. 
 

Available digits are:

• two 2's,
   
• three 3's,
   
• four 4's.
   

Since a 4-digit number cannot use more than four 4's, only the following invalid numbers were
 overcounted:

• 3222 (using three 2's when only two 2's available)
   
• 4222(using three 2's when only two 2's available)
   
• 3333(using four 3's when only three 3's available)
  
   

Hence,

54 − 3 = 51

Comments

Great explanation !!

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mind blown!

Answer 2

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

Comments

if you used all the 3's and all the 4's in the first place you will end up forming the same number over and over again, and the question was to count the distinct numbers that can be formed. Instead you should use one 3 and one 4 to get distinct numbers.

For instance, suppose I used the first 3 for the first place and the first 4 for the second place, then the 2 then the 2, I'll end up with the number 3422 right? And if I used the second 3 and the second 4 then 2 then 2, I'll also end up with the same number 3422. Which we don't want to count it another time. (The keyword is Distinct).

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@Arjun sir , i used the formula  of permutation with similar objects https://www.ck12.org/probability/permutations-with-repetition/lesson/Permutations-with-Repetition-BSC-PST/#:~:text=There%20is%20a%20subset%20of,of%20objects%20that%20are%20identical.

,But answer i get is wrong , can you please explain where i am going wrong

here is my approach  

case 1:fix 4 at msb = 8*7*6/2!3!3! [ 2! for 2s repetition, 3! for 3 3s , 3! for 3 repitions of 4s as one 4 is fixed ]

case 2:fix 3 at msb = 8*7*6/2!4!2! [2! for two 2s , 2!for remaining two 3s after fixing one 3 at msb , 4! for 4 no of 4s]

total answer = 3.5 + 4.66

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@14ash   in the question of MISSISSIPPI  they have asked possible no of arrangement with no repetition , in the above quetion they have specified that out of given digits we have to form a four-digit number in that also they told the number should be greater than 3000….

in the question of MISSISSIPPI, they have not to ask you to form a four-letter with four letters and any other condtions … they have just asked to arrange that word so, obivously as letter MISSISSIPPI have 11 words we will only have to find letters which can form using 11 words ,and for distinctness of the words  we wil divide total no of arrangements possible with products of  number of time the words repeated in that letter …...i.e.           11 ! /2!.4!..4!

Answer 3

For the case of the Number greater than 3000
Let us make the choice form where :-

1^st Position has 2 choices to make from (3 or 4)

2^nd Position has 3 choices to make form (2 or 3 or 4)

3^rd Position has 3 choices to make form (2 or 3 or 4)

4^th Position has 3 choices to make form (2 or 3 or 4)

 

Hence total of 2*3*3*3 = 54 Choices 
But here we have done a mistake as it can include number such as 4222, 3222 and 3333. So we need to subtract these. 

Hence net resultant will be = 54-3 = 51 Choices. 

Answer 4

Greater than 3000
⇒ First digit: 3 or 4
(i) First digit - 3:
We have to choose 3 digits from (2, 2, 3, 3, 4, 4, 4, 4).
Any place can have either 2 or 3 or 4, but (222, 333) is not possible as we have only two 2's and two 3's.
Total = 3 × 3 × 3 - 2 = 25
(ii) First digit - 4:
We have to choose 4 digits from (2, 2, 3, 3, 4, 4, 4, 4).
Any place can have either 2 or 3 or 4, but (222) is not possible we have only two 2's.
Total = 3 × 3 × 3 - 1 = 26
∴ Total number possible = 25 + 26 = 51

Comments

Find the number of straight lines obtained by joining N points on a plane, if No three of which are collinear .What will be the number if p points out of n are in same str line?
 

Meaning of last line ? if n=8 ,p=4 then what ?

Answer 5

_ _ _ _

Left Most Spot (LMS) can have either 3 or 4 for a number to be greater than 3000

Case 1: Leftmost is 4

SS = {2,2,3,3,3,4,4,4}

The remaining 3 spots can be filled –

_ _ _ _

For green marked spots

Leftmost → taken by 4 → remaining 2 spots can be filled in 9 ways {2,3,4}

Leftmost → taken by 3 → remaining 2 can be filed in 9 ways {2,3,4}

Leftmost → taken by 2 → Middle spot taken by 2 → Last spot can be filled in 2 ways by {3,4}

                                          Middle spot taken by {3,4}  → Last spot can be taken in 3 ways twice by {2,3,4}

Total cases = 9 + 9 + 2 + 2(3) = 26

Case 2: Leftmost is 3

_ _ _ _

SS = {2,2,3,3,4,4,4,4}

_ _ _ _

Leftmost → taken by 4 → Remaining 2 spots can be filled in 9 ways

Leftmost → taken by 3 → Middle spot taken by 3 → Last spot can be taken in 2 ways

                                     → Middle spot taken by {2,4} → Last spot can be taken in 3 ways twice by {2,3,4}

Leftmost → taken by 2 → Middle spot taken by 2 → Last spot can be taken in 2 ways

                                      → Middle spot taken by {3,4} → Last spot can be taken in 3 ways twice by {2,3,4}

Total numbers = 9 + 2 + 6 + 2 + 6 = 25

→ Total distinct numbers = 25 + 26 = 51