Question

Problem 3. How many arrangements of MATHEMATICS are there that have ALL of the following properties:

TH appear together in this order and E appears somewhere before C?

Answer 1

The word "MATHEMATICS" has 2M, 2A, T, TH, E, C, I, S.
We will consider TH as 1 Character so, we need to make arrangements for 10 characters.

_ _ _ _ _ _ _ _ _ _
(1) When E occupies the 9th place, then C can occupy only one position at 10th.
Total Arrangements = 8!/(2! * 2!)
(2) When E occupies the 8th place, then C can occupy 2 position at 9th and 10th.
Total Arrangements = 8!/(2! * 2!) * 2
(3) When E occupies the 7th place, then C can occupy 3 position at 8th, 9th and 10th.
Total Arrangements = 8!/(2! * 2!) * 3
(4) When E occupies the 6th place, then C can occupy 4 position at 7th,8th, 9th and 10th.
Total Arrangements = 8!/(2! * 2!) * 4
(5) When E occupies the 5th place, then C can occupy 5 position at 6th, 7th, 8th, 9th and 10th.
Total Arrangements = 8!/(2! * 2!) * 5
(6) When E occupies the 4th place, then C can occupy 6 position at 5th, 6th, 7th, 8th, 9th and 10th.
Total Arrangements = 8!/(2! * 2!) * 6
(7) When E occupies the 3rd place, then C can occupy 7 position at 4th, 5th, 6th, 7th, 8th, 9th and 10th.
Total Arrangements = 8!/(2! * 2!) * 7
(8) When E occupies the 2nd place, then C can occupy 8 position at 3rd, 4th, 5th, 6th, 7th, 8th, 9th and 10th.
Total Arrangements = 8!/(2! * 2!) * 8
(9) When E occupies the 1st place, then C can occupy 9 position at 2nd, 3rd, 4th, 5th, 6th, 7th, 8th, 9th and 10th.
Total Arrangements = 8!/(2! * 2!) * 9


Total Arrangements = [8!/(2!*2!)]*(1+2+3+4+5+6+7+8+9) = 8*7*6*5*4*3*2*1*45/4 = 453600 arrangements