Latest Infosys Aptitude Question SOLUTION: Q. HOW MANY 4 DIGIT NUMBER CAN BE FORMED SUCH THAT IT CONTAINS EXACTLY 3 DIFFERENT DIGITS ??

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Q. HOW MANY 4 DIGIT NUMBER CAN BE FORMED SUCH THAT IT CONTAINS EXACTLY 3 DIFFERENT DIGITS ??

Asked In Infosys prabhat (10 years ago)

Solved Preetham Read Solution (13)

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Read Solution (13) :

Preetham  (10 years ago)

Case I:
The repeated digit is the unit's digit.
So, the 1st, 2nd and 3rd digits can be selected in 9 x 9 x 8 ways, respectively.
Now the 4th digit (unit's digit) can be either equal to the 1st, 2nd or 3rd digit.
Thus, in all we have:
9x9x8x3

Case II:
The repeated digit is the ten's digit.
So, the 1st, 2nd and 4th digits can be selected in 9 x 9 x 8 ways, respectively.
Now the 3rd digit (ten's digit) can be either equal to the 1st or 2nd digit.
Thus, in all we have:
9x9x2x8

Case III:
The repeated digit is the hundred's digit.
So, the 1st, 3rd and 4th digits can be selected in 9 x 9 x 8 ways, respectively.
Now the 2nd digit (hundred's digit) is equal to the 1st digit.
Thus, in all we have:
9x1x9x8

In totality, we have 9x9x8(3+2+1) = 9x9x8x6 = 3888

Hope this helps.

Is this solution Helpfull? Yes (34) | No (11)  

RAKESH  (10 years ago)

4 DIGIT NUMBER WITH EXACTLY 3 DIFFERENT DIGITS i.e repition not allowed
total no. of digits =10(0,1,2,3,...,9)
thousand place cant be 0 so can be filled in 9 ways
rest three places can be filled in 9P3 ways
total = 9* 9P3 = 9*9*8*7= 4536

Is this solution Helpfull? Yes (18) | No (11)  

shriya maini  (10 years ago)

answer is 1944 (9.9.8.3=1944)
1st digit can be selected in 9 ways
2nd in 9 ways again
3rd in 8 ways. This way we have 3 distinct digits.
now the 4th digit has to be a repeat of 1st digit or 2nd or 3rd digit
Hence it can be done in 3 ways
So, 9.9.8.3=1944

Is this solution Helpfull? Yes (17) | No (2)  

kriti yadav  (10 years ago)

ans is 3888
no of arrangement=3!=6
one digit should be repeated,n no of possible arrangement is 6
so 9*9*8*6=3888

Is this solution Helpfull? Yes (4) | No   

RAKESH  (10 years ago)

asking for 3 difffer digit
its can be formed by 2 same & 2 differ digits
11,22,...,99 => 9(4!/2! *9c2 - 3!/2! *8)
00 => 9c1*8c1*(3!/2!)
total=9(4!/2! *9c2 - 3!/2! *8)+9c1*8c1*(3!/2!)

is it right??

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Satyaranjan  (10 years ago)

@K Arun Ya u r right.3888 is the right answer.bcoz it is asking for exactly 3 diff digits (not exactly 4 diff digits):
So answer will be 8424(from my previous calculation)-4536(nos with exactly 4 diff digits)=3888

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K.ARun  (10 years ago)

@Rakesh: Me too thought that only... but 4536 is not in the option..

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Satyaranjan  (10 years ago)

@rakesh 1134 is also a no with 3 diff digits but 1 repeated

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Satyaranjan  (10 years ago)

1.Four digit nos with no digits repeated:9x9x8x7=4536
2.Nos with two zeros(ex:1004): (9C1x8C1x3!)/2!=9x8x3=216
3.Nos with two 1's:(ex:1123,2311,as these have exactly three different digits 1,2 and 3)
i.nos starting with 1(ex:1123):9C2x3!=216
ii.nos not starting with 1(ex:2311):(8C1x8C1X3!)/2!=192
So total nos with two 1's=216+192=408
Similarly for two 2's=408,two 3's=408...
so total nos with exactly three different digits=4536+216+9x408=8424
@K Arun Is this option present there?

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K.ARun  (10 years ago)

@satyaranjan: No,
options are
3280
4216
3888
Ans might be 3888 not sure

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surbhi   (10 years ago)

@k arun could u plz upload more ques?any ques repeated from here?

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Satyaranjan  (10 years ago)

@K.ARUN :it is impossible because there will be 4536 no if it contains exactly 4 different digits.

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T G  (10 years ago)

9*9*8*3=1944? is there any option like this?

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