A probability problem by Abdullah Zafar

Probability Level 2

How many 5-digit numbers greater than 23000 that can be formed from digits 1,2,3,5,6 without repeating any digit?

for example, 23156 is a number greater than 23000.

380 90 45 176

Abdullah Zafar by Abdullah Zafar , 22, Pakistan

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3 solutions

Mukul Rathi
Dec 24, 2014

The total number of such numbers is 5!=120 as there are 5 options for the first digit, 4 for the second..... ...1 for the last = 5x4x3x2x1 = 5! = 120 (For those unfamiliar with factorial notation, it is probably best that you check that out as I will be using that quite a lot - simply put n!= n x (n-1) x ... ...3 x 2 x 1.

From this we must subtract the numbers that start with a 1. 1 is fixed so permuting the other 4 digits we get 4! = 24

We must also subtract the numbers that start with 21. 2 and 1 are fixed so again permuting the other digits we have 3!=6

So the total of such numbers is 120 - 24 -6 = 90.

2 Helpful 0 Interesting 0 Brilliant 0 Confused
Cp Gupta
Dec 22, 2014

2####=4!=24 but 21###=3!=6 So 24-6=18then 3####=4!=24 5####=4!=24 6####=4!=24 So 18+24+24+24=90

0 Helpful 0 Interesting 0 Brilliant 0 Confused
Abdullah Zafar
Dec 14, 2014

no.s of form 23$$$=3!=6

no.s of form 25$$$=3!=6

no.s of form 26$$$=3!=6

no.s of form 3$$$$=4!=24

no.s of form 5$$$$=4!=24

no.s of form 6$$$$=4!=24

total no.s=6+6+6+24+24+24=90 So, 90 is correct answer.

0 Helpful 0 Interesting 0 Brilliant 0 Confused

All numbers greater than 20000 is 24x4=96

no.s of form 21$$$=3!=6

96-6=90

BK Lim - 6 years, 6 months ago

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