A Person Starts Writing All 4 Digits’ Numbers. How Many Times Had H

1.	A Person Starts Writing All 4 Digits’ Numbers. How Many Times Had He Written The Digit 2?
Answer» Number of 2‘s at unit’s place (from 100-1 to 999-2) = 900 Number of 2‘s at tenths place (xy2z) xy will vary from 10 – 99 (90 NOS) and z from 0-9 (10 nos) so TOTAL 9010 = 900 Number of 2‘s at HUNDREDS place(x2yz) X will vary from 1-9(9 ways) & yz from 00-99(100 ways) so total 9100 =900 Number of 2‘s at thousands place(2xyz) xyz will vary from 000-999 so total = 1000 Therefore, total number of 2‘s = (900+900+900+1000) =3700. Number of 2‘s at unit’s place (from 100-1 to 999-2) = 900 Number of 2‘s at tenths place (xy2z) xy will vary from 10 – 99 (90 nos) and z from 0-9 (10 nos) so total 9010 = 900 Number of 2‘s at hundreds place(x2yz) x will vary from 1-9(9 ways) & yz from 00-99(100 ways) so total 9100 =900 Number of 2‘s at thousands place(2xyz) xyz will vary from 000-999 so total = 1000 Therefore, total number of 2‘s = (900+900+900+1000) =3700.

A Person Starts Writing All 4 Digits’ Numbers. How Many Times Had He Written The Digit 2?

Answer»

Number of 2‘s at unit’s place (from 100-1 to 999-2) = 900

Number of 2‘s at tenths place (xy2z) xy will vary from 10 – 99 (90 NOS) and z from 0-9 (10 nos) so TOTAL 90*10 = 900

Number of 2‘s at HUNDREDS place(x2yz) X will vary from 1-9(9 ways) & yz from 00-99(100 ways) so total 9*100 =900

Number of 2‘s at thousands place(2xyz) xyz will vary from 000-999 so total = 1000

Therefore, total number of 2‘s = (900+900+900+1000) =3700.

Number of 2‘s at unit’s place (from 100-1 to 999-2) = 900

Number of 2‘s at tenths place (xy2z) xy will vary from 10 – 99 (90 nos) and z from 0-9 (10 nos) so total 90*10 = 900

Number of 2‘s at hundreds place(x2yz) x will vary from 1-9(9 ways) & yz from 00-99(100 ways) so total 9*100 =900

Number of 2‘s at thousands place(2xyz) xyz will vary from 000-999 so total = 1000

Therefore, total number of 2‘s = (900+900+900+1000) =3700.