What is the sum of all the possible 3 digit numbers that can be constructed using the digits 3,4,and 5 if each digit can be used only once in each number?
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@ Kapil
Your md is simpler fr this sum.. But this formula enhances our capability ot solvesuch sums even with more no of digits… and generic formulae always help .. thanks atul for the formula and explanation.. very helpful
is this method simple or formula
the possible no. of 3,4,5 are
345
354
435
453
534
543
____
2664
____
@ Atul
Great Explanation !!! Thanks!!
@vertica,
Apologies for the delay response .. Yes this method is applicable for all the scenarios.
@Charanya / manisha,
Nice formula .. there is a slight change is the formula to make it generic:
suppose n is the no. of digits given. Our scenario it is 3 (digits given 3,4,5)
Formula will be:
(n times 1) * {fact (n-1)} * (sum of digits)
explaination:
First term : ‘n times 1′
for n = 3 this means ’3 times 1′ i.e. 111
for n = 4 this will be ’4 times 1′ i.e. 1111
Second term : fact(n-1) ……..(factorial of (n-1)
for n = 3 this means fact(3-1) = fact (2) = 2
for n = 4 it will be !(4-1) = !3 = 6
Third term : Sum of the digits
in our case = 3+4+5 = 12
so the solution of our problem with this formula will be as Charanya explained:
= (3 times 1) (fact(3-1) (3+4+5)
= (111)(!2)(12) = 111*2*12
= 2664
This formula can be used in the scenarios were ’0′ is also present in the given digits. Hope it will help.
@manisha…
its a simple formula which someone taught me …
(111..n times)(n – 1) (Sum of digits)
The formula varies when one of the digits is zero.
I’ll post it ASAP.
Hey Atul
Can this method be used for any kind of problem similar to give here?
Nice concept..do you have some more tricks/ logical solutions…i would appreciate if you would share it!
@ Charanya
What is the basic concept behind this formula? whr all it can be actually used?
there is a simple formula for it –
(111)(n-1)(sum of the digits)
= (111)(2)(12)
= 2664
Since each digit can appear only once.make the combinations by fixing one no and interchanging the other two .the nos are :
345,354
435,453
534,543
Now add the above nos .The result is 2664
Total should be = 345+354+435+453+543+534=2664
To form the 3 digit number using 3,4,5, total no. of ways to form a number with the given number without repetation :
3*2*1 = 6
Let us suppose three digit nol is ABC (A at hundreth place, B tense place, C at Unit place)
when we form different combinations of the 3 digit numbers. No. of time every no. can appear on a particualar place (Unit, Ten etc) is
6 (Total possible combinations) /3 (Total digits available) = 2
So every no. will come 2 times in all the places (’3′ will come 2 times at A, 2 times at B and 2 times at C in total combinations. similarly 4 and 5). So sum total for ”Unit’s place” Ten’s place and Hunderd’s place individually are:
Unit’s place = 3*2 + 4*2 + 5*2 = 24
Ten’s place = 4*2 + 5*2 + 3*2 = 24
Hundred’s place = 5*2 + 3*2 + 4*2 = 24
forming the final total:
24 / 24 / 24
= 24 / (24 + 2) / 4 (Carry fwd 2 from unit’s place to ten’s place)
= 24 / 26 / 4
= (24 + 2) / 6 / 4 (Carry fwd 2 from ten’s place to Hunderd’s place)
= 26/6/4
Final answer = 2664
The method will be very effective for higher digit nos (example sum of 5 digit no. using 1,2,3,4 and 5).
6 different combination.
2664
thats right answer!!!!
345 + 354 + 435 + 453 + 543 + 534 = 2664
1.B
2.C
3.E
4.C
5.E
6.C
2664
Though Options are not given….but according to me….
Possible 3 Digit Nos. are: 345, 354, 435, 453, 543, 534 and hence the sum of these nos. is = 2664