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

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.

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MY first approach was 7*8*7*6=2352 ways .
WHat is wrong in this apporach ? Am i not excluding the repeated parts?

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We considered selections of ${\color{Red}{\textbf{k items}} }$ out of ${\color{Magenta}{\textbf{n possible options.}} }$

	$\textbf{With repetitions}$	$\textbf{Without repetitions}$
$\textbf{Ordred}$	$\text{Tuples}:n^{k}$	$k-\text{permutations}:\dfrac{n!}{(n-k)!}$
$\textbf{Unordered}$	$\text{Combinations with repetitions}:\binom{k + n - 1}{n-1} = \binom{k + n - 1}{k} $ Also known as Integer Equations - Stars and Bars	$\text{Combinations}:\binom{n}{k}$

Answer 1

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

Answer 2

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 3

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 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 ?