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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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ManjariMishra
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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You are right. The question should mention that we are looking for 5-digit numbers only.
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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.
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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) \(120\)

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.
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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
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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?
KarishmaB
sharathnair14
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) \(120\)

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.
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