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Question Number 218129 by mr W last updated on 30/Mar/25

h o w m a n y d i f f e r e n t w o r d s c a n b e

f o r m e d f r o m t h e w o r d

MATHEMATICS ?

n o t e : h e r e a w o r d s h o u l d h a v e a t

l e a s t t w o l e t t e r s, b u t {m u s t n}^{'} t h a v e a

m e a n i n g .

Answered by vnm last updated on 30/Mar/25

\underset{n = 2}{\sum^{8}} C_{8}^{n} n! + 3 \underset{n = 0}{\sum^{7}} C_{7}^{n} n! C_{n + 2}^{2} +

3 \underset{n = 0}{\sum^{6}} C_{6}^{n} n! C_{n + 4}^{2} C_{n + 2}^{2} + \underset{n = 0}{\sum^{5}} C_{5}^{n} n! C_{n + 6}^{2} C_{n + 4}^{2} C_{n + 2}^{2} =

= 13938212

t h e s a m e r e s u l t :

\underset{n = 2}{\sum^{11}} \underset{k = 0}{\sum^{\min (3, [\frac{n}{2}])}} C_{8 - k}^{n - 2 k} (n - 2 k)! C_{3}^{k} \underset{i = 0}{\prod^{k - 1}} C_{n - 2 i}^{2}

Commented by mr W last updated on 31/Mar/25

g r e a t! t h a n k s s i r!

Answered by mr W last updated on 31/Mar/25

M, A, T c a n b e u s e d a t m o s t t w o

t i m e s e a c h .

H, E, I, C, S c a n b e u s e d a t m o s t o n e

t i m e e a c h .

{(1 + x + \frac{x^{2}}{2!})}^{3} {(1 + x)}^{5}

= 1 + 8 x + \frac{59 x^{2}}{2} + \frac{133 x^{3}}{2} + \frac{409 x^{4}}{4} + 113 x^{5}

+ \frac{735 x^{6}}{6} + \frac{441 x^{7}}{8} + 24 x^{8} + \frac{29 x^{9}}{4}

+ 11 x^{10} + \frac{x^{11}}{8}

t o t a l n u m b e r o f d i f f e r e n t w o r d s

w i t h 2 t i l l 11 l e t t e r s :

\frac{59 \times 2!}{2} + \frac{133 \times 3!}{2} + \frac{409 \times 4!}{4} + 113 \times 5!

+ \frac{735 \times 6!}{6} + \frac{441 \times 7!}{8} + 24 \times 8! + \frac{29 \times 9!}{4}

+ 11 \times 10! + \frac{11!}{8} = 13 938 212 ✓

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