write the three digit numbers where hundred place is a tenth place is 4 and unit place is b
Answers
Answer:
There are 10 digits.. 0, 1, 2, 3, 4, 5, 6, 7, 8, 9…
Let’s start putting each digit in middle and figure out probability of other two digits..
If we consider digit 0 in middle, there is no digit which is smaller than 0.. So, this choice is dropped.
Consider digit 1 in the middle, there is only 1 digit which is smaller than 1 and that is 0… If we put 0 in the beginning.. that will not be a 3 digit number.. So, this choice is also dropped.
Consider digit 2 in the middle.. two digits are smaller than 2 : 1 and 0. Again, 0 can not be put at the 1st place as it will not make a 3 digit number. So, possible number are :-
120 and 121
Consider digit 3 in the middle… there are 3 digits small than 3 which are 0, 1 and 2 but again 0 can not be 1st digit (It will be applicable in all below cases, so i won’t mention it again..). So, possible numbers are :-
Let’s write digit 3 in the middle and all possible digits on 1st place..
13
23
Now, let’s us write all possible digits at the 3rd place for both the above digits :
130
131
132
230
231
232
Same way for digit 4.. we will have..
4 digits smaller than 4 : 0, 1, 2, 3
Write 1st 2 digits (0 at 1st place not possible) :
14
24
34
Write last digit including all smaller digits 1 by 1 at the 3rd place…
140
141
142
143
240
241
242
243
340
341
342
343
For digit 5… (using the same trick as above..)
Initial set with 1st and 2nd digit :-
15
25
35
45
Final Set with all 3 digits :-
150
151
152
153
154
250
251
252
253
254
350
351
352
353
354
450
451
452
453
454
Same way you can do for other digits and count manually…
Now, analyze it…
As per above trick… For every digit n, you 1st have (n-1) no’s with 2 digits and then add n digits to end for every number…
So, for every digit n, you actually have n*(n-1) 3 digit numbers as per requirement…
For n=0, we will have 0*(0–1) which is 0.. That is none…
For n=1, we will have 1 *(1–1) which is again 0.. That is none…
For n=2, 2*(2–1) = 2
and so on…
So, final answer to your question is…
0*(0–1) + 1*(1–1) + 2*(2–1) + 3*(3–1) + 4*(4–1) + 5*(5–1) + 6*(6–1) + 7*(7–1) + 8*(8–1) + 9*(9–1) = 0 + 0 + 2 + 6 + 12 + 20 + 30 + 42 + 56 + 72 = 240