Question

4. Counting Problems. (a). Find number of ordered triples (x, y, z) of strictly positive integers such that 2 + y +z = 111.

Answer 1

The question is solved using required formulas with detailed explanation.

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2 ga) au+y+z = 1 Total number of ordered triplets na sum of all variables ra total of variables, Now, total no. of ordered triplets casy,z) are where 110. Cogel 110 = 5995 s=15' Total alphabets) MATHEMATICS. M=2s' Tags 1 =15 E = 15' I -15 c=15 A=25' as BLON Result must contain "EC" So, we will count them 1. So total alphabets - 11-1-10 Now, no. of 10! ways to arrangje 2!2!2! 21 divided 8 times because 3628800 there are 3 letteres with repeatation 453600 ways MATHEMATICS & É muet occurs somewhere left to 'C'. So, 'd' can't be placed at the 1st place. Ele Now, if we put C at the and place, then & must

CON be done can arrange the rest of the al letters be placed at the firest place & this is only I way. Now we a! = 45360 ways 2!2!?! ways a E be placed in left of 'C in 3rd place, 50, E can in 29-2 ways, And, rest al letteres in 91 -45360 ways. 2!2!? 90720 ways Total ways 2x 45 360 F in al 121, 12345360 c'in 4th place so, 'E' can be placed is 30 - 3 was Ref lettores 1345360 ways Total ways = 3845560 -136080 ways 1) 'C' in 5th place. So t can be placed in the ways Rest 'a' letteres in all 2121 waye : 45 360 ways, Total coays - 4x45360 TQ1490.00wh - 181440000 E in 6th place, & can be placed in 5a = 5 ways, Rest a letters in 91/21/21 Totat a 5845360 226800 ways - 45360 ways.

If we go im similar trond then & For'c' in 4th plane Total ways - GX45360 - 272160 ways fore ' in 8th place Total waye 7x45360 317520 ways, for ' in oth place Total woys - 8x45 360 236 2880 ways, & For o in 10th place Total ways -9x45360 = 408240 .coays. For C in 11th place Toto) ways a 10X453602 453600 ways Now, number of ways to arrange so that E' must occure at leffside of cc is 45360 + 90720 + 135080 +181440 + 226800 +272160 +317520 +362880 +408240 tu 53600 > 2494800 waye. Cans.)