how many six digit numbers contain exactly three different digits
Answers
Step-by-step explanation:
100 approximately........
56160 are six digit numbers contain exactly three different digits
Step-by-step explanation:
To find Six digit numbers Containing exactly three Different Digit
I will give you the approach
we have 10 Digits from 0 to 9
we can Select 3 digits out of 10 in ¹⁰C₃ = 120 Ways
Numbers having 0 as one of Digit = ⁹C₂ = 36 Ways
Numbers not having 0 as one of digit = 120 - 36 = 84 Ways
Numbers not having 0 as one of digit
let say digits abc
these can be 411 , 321 , 222 ,
411 can be in 3 ways and numbers in each case = 6!/4! = 30
321 can be in 6 ways and numbers in each case = 6!/(3!*2!) = 60
222 can be in 1 way and numbers = 6!/(2!*2!*2!) = 90
total numbers = 3 * 30 + 6 * 60 + 90 = 540
84 * 540 = 45360
explained below
aaaabc - 6!/4! = 30
aaabbc - 6!/(3!*2!) = 60
aaabcc - 6!/(3!*2!) = 60
aabbbc - 6!/(3!*2!) = 60
aabccc - 6!/(3!*2!) = 60
aabbcc - 6!/(2!*2!*2!) = 90
abbbbc - 6!/4! = 30
acbbbb - 6!/4! = 30
abbbcc - 6!/(3!*2!) = 60
acccbb= - 6!/(3!*2!) = 60
Now 36 cases where 0 is one of the digit
so we have 0ab
1st digit can not be 0 hence
1st Digit 2 ways a or b
Remaining 5 digits
0ab can be
311 , 221
311 can be in 3 ways and numbers in each case = 5!/3! = 20
221 can be 3 ways and numbers in each case = 5!/(2!*2!) = 30
numbers = 2 ( 3 * 20 + 3 * 30) = 300
36 * 300 =10800
explained below
a - 1st digit & next 5 digit
aaabc - 5!/3! = 20
bbbac - 5!/3! = 20
cccab - 5!/3! = 20
aabbc - 5!/(2!*2!) = 30
aaccb - 5!/(2!*2!) = 30
bbcca - 5!/(2!*2!) = 30
b - 1st digit & next 5 digit
aaabc - 5!/3! = 20
bbbac - 5!/3! = 20
cccab - 5!/3! = 20
aabbc - 5!/(2!*2!) = 30
aaccb - 5!/(2!*2!) = 30
bbcca - 5!/(2!*2!) = 30
45360 + 10800 = 56160 are six digit numbers contain exactly three different digits
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