How many numbers can be formed from 1, 2, 3, 4, 5 (without repetition)

Question

How many five digit numbers can be formed from 1, 2, 3, 4, 5 (without repetition), when the digit at the unit’s place must be greater than that in the ten’s place?

(a)  54 
(b)  60 
(c)  17 
(d)  2 × 4! 
(e)  120

BrushMyQuant · Accepted Answer

Total Number of Numbers which can be formed by numbers 1,2,3,4,5 (without repeating digits)  = 5*4*3*2*1 = 5! = 120.
Now, in   half   of them unit's digit will be   bigger   than the ten's digit and in   half   of them it will be   smaller  .

Example  : Let's say we have three digits 1,2,3.
Total number of numbers without repeating digits = 3*2*1=6
Numbers with Unit's digit greater than the ten's digit
123, 213, 312
Numbers with Ten's digit greater than the unit's digit
321, 132, 231

So,  Total Number of cases = 120/2 = 60

So,  Answer will be B 
Hope it helps!

jackfr2 · Answer

unit's place>ten's place

So , possible unit digit = 2.3.4.5

when 2 is in unit's digit 1 must be in ten's and (3,4,5) forms the other numbers.

total possible number =3!=6

similarly when 3 is in unit's digit 1 or 2 can be in ten's digit and 3 other digits form the number.

so total possible number =3!*2=12

again when 4 ................. total possible number =3!*3=18

and when 5 .................. total possible number =3!*4=24

sum of total possibilities =6+12+18+24=60

Answer: B

ManjariMishra · Answer

Does it mean a five digit number? A number can be 2 digit , 3 digit till 5 digit for this combination

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Bunuel · Answer

You are right. The question should mention that we are looking for 5-digit numbers only.

gvij2017 · Answer

Condi-1:Digit at unit place> digit at tens place.
Condi-2: Without repetition
(1,2,3,4,5)

possible combinations for tens place and unit place, 5C2= 10. Here we will not multiply by 2! because we want ascending order. For example, (2,1) and (1, 2) are two pair but we need only (2,1) which is satisfying condition-1

For remaining places, arrangement of remaining digits is 3*2*1= 6. 
So total ways of arrangement= 6*10= 60. 
B is answer.

KarishmaB · Answer

No of 5 digit numbers with 1, 2, 3, 4, 5 digits = 5! = 120

By symmetry, in half of them, the units digit will be greater that tens digit and in the other half, the tens digit will be greater than units digit.

So 120/2 = 60

Answer (B)

Note the symmetry - If 1 is in units digit, all such numbers will not be included. If 5 is in the units digit, all such numbers will be included. If 2 is in units digit, only numbers with 1 is tens digit will be included. If 4 is in units digit, only number with 5 in tens digit will not be included. When 3 is in units digit, half the numbers will be acceptable and half will not be.

Paras96 · Answer

How many five digit numbers can be formed from 1, 2, 3, 4, 5 (without repetition), when the digit at the unit’s place must be greater than that in the ten’s place?

We will consider pairs for the digit at the unit’s place and that in the ten’s place,
(2,1)(3,1)(4,1)(5,1)(3,2)(4,2)(5,2)(4,3)(5,3)(5,4) = 10

The other 3 places can be arranged in 3! ways i.e. 6 ways

Therefore required ways = 6x10=60 ways

Hence B

kanwar08 · Answer

Is there any general rule on symmetry? What if it were a 6-digit integer with 1,2,3,4,5,6(w/o repetition).

Will this rule hold true in this case? i.e. 6!/2?

How many numbers can be formed from 1, 2, 3, 4, 5 (without repetition)

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How many numbers can be formed from 1, 2, 3, 4, 5 (without repetition)

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