A problem
Three boys randomly drew 4 digits from a bag having a large
collection of numbers from 0 to 9. Boy A got 4 different digits without zero,
boy B had 3 different digits with one zero while boy C was given with 3
different digits and one identical digit. How many different numbers each one
of them can display with their digits?
Boy A got 4 different digits [1,2,3,4]
Single digit display
4 [1,2,3,4] = 4C1
Double digits display. Fix a digit as first digit, then
choose any one out of three. Repeat this for other digits. 4 [3C1]
=12 { 1(,2,3,4)-3;2(1,3,4)-3; 3(1,2,4)-3
; 4(1,2,3)-3}
Triple digits display. Fix first digit. then choose any one
out of three and then choose any one out of two.
4[3C1 x 2C1] = 4
x 3 x 2 = 24 e.g., with 1 as first digit there will be 6
numbers 1 {23,32,34,43,42,24}
For four digit display
4C1 [3C1 x2C1
x2C1 x1C1 ] = 24 for example with 1 as first digit there will
be 6 numbers 1 { 234,243,324,342,423,432}
Hece Boy A can show 4+12+24+24 = 64 numbers
Boy B has 3 different
digits [1,2,3] and one zero [0]. Zero cannot exist in the beginning of any
number.
Single digit display 3 (1,2,3]= 3C1
Double digits display. Choose one digit out of three and
then choose any one out of the remaining three
3 [3C1] – 9 e.g., with one as first digit 1 [0,2,3]- 3
Triple digits numbers
3 [ 3C1 x 2C1 ] =3x3x2 = 18
Four digits numbers 3 [3C1x2C1x1C1
] = 3x3x2 = 18
B can display 3 +9+18+18 = 48 numbers
Boy C has 3 different numbers [1,2,3] and one identical
number [1].
Single digit number 3C1
= 3 [1,2,3]
Double digits numbers
for 1 3C1 and for 2,3 – 2 x2C1
= 3 +2x2 = 7
Triple digits numbers
3 [3C1 x 2C1 ] = 18 numbers
Four digit numbers 3 [3C1x2C1x1C1
]= 18 numbers
Boy C can display 3+7+18+18 = 46 numbers
