code is a sequence of 5 digits and 2 letters(ex123a45b) find the prob. code starts w/9 & ends w/z. wout/rep?

Answer 1

first, state a question without causing confusion....avoiding needless abbreviations...!

since there is no condition for digits and letters, they can appear any way... so we consider them together... 7 characters in all...so 7! possible ways of arranging them... the five digits can be chosen in 10C5 ways and 2 letters in 26C2 ways...

so total combinations: 10!26!7! / 5!5!24!2! = 10.9.8.7.26.25.7! / 5.4.2 = 126.650.7!

once we assume a code starts with 9 and ends with z, we can pick remaining letters in 25 ways and digits in 10C4 ways; their permutation will be 5!

so possible ways: 25*10*9*8*7*5!/24

probability = 25.10.9.8.7.5!/24.126.650.7! = 10.9.8.7 / 24.26.6.7
=5/26

(the method is as indicated... i did not recheck calculations)

Answer 2

1) When starts with word and ends with word
The first and last place is filled by 1x1=1ways.......i
Now five places are left to fill with 26+10=36
36C5=376992 ways.............................ii
Total number of ways 0f selection= 1x376992
Probability=376992/38C7=0.029871977=0.03 app

Answer 3

95a732z

Answer 4

That would be about 0.00067022823%