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sharathnair14
Joined: 05 May 2019
Last visit: 27 May 2020
Posts: 163
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Schools: LBS '25 ISB '25 Manchester'26 Stern '26 Tuck '26 INSEAD Aug'24 intake Yale '26 Kellogg '26 Wharton '26
GPA: 3
Schools: LBS '25 ISB '25 Manchester'26 Stern '26 Tuck '26 INSEAD Aug'24 intake Yale '26 Kellogg '26 Wharton '26
Posts: 163
Updated on: 20 Jan 2020, 06:37
Originally posted by sharathnair14 on 10 Jan 2020, 10:38.
Last edited by sharathnair14 on 20 Jan 2020, 06:37, edited 1 time in total.
Last edited by sharathnair14 on 20 Jan 2020, 06:37, edited 1 time in total.
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
70% (02:12) correct
30%
(02:13)
wrong
based on 479
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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) \(120\)
(a) \(54\)
(b) \(60\)
(c) \(17\)
(d) \(2 × 4!\)
(e) \(120\)
Updated on: 20 Jul 2024, 02:53
Originally posted by BrushMyQuant on 11 Jan 2020, 09:59.
Last edited by BrushMyQuant on 20 Jul 2024, 02:53, edited 2 times in total.
Last edited by BrushMyQuant on 20 Jul 2024, 02:53, edited 2 times in total.
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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!
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!
19 Jan 2020, 02:24
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sharathnair14
How many 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) \(120\)
(a) \(54\)
(b) \(60\)
(c) \(17\)
(d) \(2 × 4!\)
(e) \(120\)
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
