How many numbers of five digits can be formed by using 1, 2, 3, 4, 5, if 4 is necessarily taken at

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25 Nov 2009 five digits number

How many numbers of five digits can be formed by using 1, 2, 3, 4, 5, if 4 is necessarily taken at hundred's
place and 2 is not allowed at unit's place. Repetition of numbers is not allowed.

OPtion

1) 72
2) 16
3) 12
4) 120
5) 96
6) 24
7) 60
8) 30
9) 15
10) 18

Solution

4 is necessarily taken at hundred's place and 2 is not allowed at unit's place,
hence number of methods = 1*3*3*2*1 = 18
You can also check the answer by making different numbers under given conditions.

Correct Option: 10 Best Solution (7)

Tijo Jose   15 years AGO

Num is ABCDE. D=4. So, 1 option. Remaining 4 digits and 4 places. E not= 2. So, 3 options. Remaining 3 digits and 3 places. A, B, C can be any one of these remaining digits. So, number of arrangements in A, B, C is 3P3 = 3!/(3-3)! = 6 options. Total number of options = 6*1*3 = 18

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susmit   15 years AGO

3*2*1*1*3 =18

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Mehul MKDP   15 years AGO

4 FACT - (2 NOT AT UNIT)

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Dr.Dipin Singh   15 years AGO

Only 4 positions are variable.

Also 2 can not come in unit's place.
so


4!-3!=18

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Sukham   15 years AGO

3*3*2*1=18

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sujit   15 years AGO

Keeping 4 fixed at 100th position.
so there are 4 digits left.
Out of which 2 can not be placed in units place.
permutation of units place is 3.
permutation of 10000th place is 3
permutation of 1000th place is 2 .
permutation of 100 place is 1
permutation of 10th place = 1
So total permutation - 3 X 2 X 3 X 1 X 1= 18

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SURAJ   15 years AGO

Total no.of digits keeping 4 at hundred's place = 4 * 3 * 2 * 1 = 24

Total no.of digits keeping 4 at hundred's place & 2 at one's place = 3 * 2 * 1 = 6

Total no.of digits keeping 4 at hundred's place & 2 not at one's place = 24 - 6 = 18

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