Latest TCS Aptitude Question SOLUTION: How many positive numbers not greater than 4300 can be formed using the digits 0,1,2,3,4 where repeatation is allowed? Options: a)576 b)575 c)620 d)640

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How many positive numbers not greater than 4300 can be formed using the digits 0,1,2,3,4 where repeatation is allowed?

Options: a)576 b)575 c)620 d)640

Asked In TCS milan jena (8 years ago)

Solved Siddhant Sengar Read Solution (11)

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Read Solution (11) :

Siddhant Sengar  (8 years ago) Best Solution

a) 576

1 digit = 5
2 digits = 4* 5 = 20
3 digits = 4*5*5= 100

Now,4 digit number till 4000 can have digits 1,2,3 in thousands place ( total choices = 3 )
0,1,2,3,4 in hundreds ,tens and ones place
I.e 3*5*5*5 = 375

Now From 4000 to 4299 = 1*3*5*5 =75


Total = 5+20+100+375+75+1 =576 (1 for 4300)

Is this solution Helpfull? Yes (41) | No (10)  

naman jha  (8 years ago)

When repitition is allowed

1 digit = 5
2 digits = 4* 5 = 20
3 digits = 4*5*5= 100

Now,4 digit number till 4000 can have digits 1,2,3 in thousands place ( total choices = 3 )
0,1,2,3,4 in hundreds ,tens and ones place
I.e 3*5*5*5 = 375

Now From 4000 to 4300 = 1*3*5*5 =75


Total = 5+20+100+375+75 =575

Is this solution Helpfull? Yes (8) | No (4)  

TAMIL  (8 years ago)

4300
1,step o*5^0=0
2,step 0*5^1=0
3,step 3*5^2=75
4,step 4*5^3=500
some 0+0+75+500=575.
ans 575

Is this solution Helpfull? Yes (8) | No   

Rohan Dey  (8 years ago)

0 is not a positive number.. the answer would be 575

Is this solution Helpfull? Yes (5) | No   

sangeetha jayasankar  (8 years ago)

Answer is 575

Is this solution Helpfull? Yes (2) | No (5)  

Deepak kumar  (8 years ago)

(4c1 * 5c1 * 5c1 * 5c1) + (1c1 * 3c1 * 5c1 * 5c1 ) + (1c1 *1c1 *1c1 *1c1 )= 500+75+1=576

Is this solution Helpfull? Yes (2) | No (3)  

Gnana Laxmi . V  (8 years ago)

b) 575

1 digit = 5
2 digits = 4* 5 = 20
3 digits = 4*5*5= 100

Now,4 digit number till 4000 can have digits 1,2,3 in thousands place ( total choices = 3 )
0,1,2,3,4 in hundreds ,tens and ones place
I.e 3*5*5*5 = 375

Now From 4000 to 4300 = 1*3*5*5 =75


Total = 5+20+100+375+75 =575

Is this solution Helpfull? Yes (1) | No (4)  

Ashish singh  (8 years ago)

first count for less than 3400.....we get 4x5x5x5 =500

now from 4000 to 4299....we get 1x3x5x5 = 75

now 4300 is left which we can write in 1 way =1

adding above 500+75+1 =576

Is this solution Helpfull? Yes (1) | No (2)  

Sitanshu  (8 years ago)

kaun best solution ka muhar maarta hai bhai kch dekh lia kar

Is this solution Helpfull? Yes (1) | No (2)  

lalitha nagarajan  (8 years ago)

One easy method is to notice that there is a one-to-one correspondence between the numbers less than 4300 that can be made from the digits 0–4 and all the numbers less than 43005 in their base-5 representation. The answer is therefore 43005=4×5^3+3×5^2=575.

Is this solution Helpfull? Yes | No   

Subhrant Pandey  (7 years ago)

4300=4×5^3+3×5^2=575.

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