Let N denote the number of ways in which 135 can be written as sum of ‘n’ natural numbers in the following manner.
$135 = a_{1} + a_{2} + a_{3} + ....... + a_{n} (n \in N)$
The digits `1, 2, 3…….9 are used to form numbers $a_{1}, a_{2}, a_{3}, ......., a_{n}$ such that all the digits are used and each digit is used exactly once in the sum. For example
$135 = \frac{18 + 95 + 2 + 3 + 4 + 6 + 7}{e a c h d i g i t f r o m 1 - 9 i s u s e d o n c e}$
Then $\frac{N}{6!} e q u a l s$

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

answer is 1676.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

None of $a_{1}, a_{2}, a_{3}, ....... a_{n}$ can be a 3-digit numbers
Hence, a few of them are 2-digit numbers, and rest are 1-digit numbers
Let sum of digits in terms in the place = k. Then sum of units place
= 45 – k
$∴ 135 = 10 \times k + 45 - k \Rightarrow 9 k = 135 - 45 = 90 \Rightarrow k = 10 k = 1 + 9 = 2 + 8 = 3 + 7 = 4 + 6 \to 4 \times^{7} P_{2} \times 7! k = 1 + 2 + 7 = 1 + 3 + 6 = 1 + 4 + 5 = 2 + 3 + 5 \to 4 \times^{6} P_{3} \times 6 k = 1 + 2 + 3 + 4 \to 1 \times^{5} P_{4} \times 5 ∴ t o t a l = 4 \times 7 \times 6 \times 7 + 4 \times 6 \times 5 \times 4 \times 6 + 5 \times 4 \times 3 \times 2 \times 5 = 21 \times 8 + 500 \times 6 = 1676 \times 6$