The number of seven digit integer is to be formed using only 1, 2 &3 such that the sum of all the digits is 10 . so how many such seven digit number is possible ?

Question

The number of seven digit integer is to be formed using only 1, 2 &3 such that the sum of all the digits is 10  . so how many such seven digit number is possible ?

Eli Assouline · Accepted Answer

I got #42# different numbers.

We start by listing the combinations of numbers that give us a digit sum of #10#. We can have #1, 1, 1, 1, 1, 3, 2# And the number of different arrangements here is #(7!)/(5!) = 42#. Is the above sequence the only possible? Note that you need a certain number of #1#'s to make the number of digits #7# and the sum #10#, because the number #2222222# has a digit sum of #14#, for example. If we try other sequences, such as #1, 1, 1, 1, 1, 1, 1, 3# We either get sequences that are more or less than #7# terms or that have a sum other than #10#. I'm not sure how to prove that the above sequence is the only one possible, so I'll leave that to other contributors. Hopefully this helps!

Jace Anderson · Answer

undefined To find the number of seven-digit integers using only 1, 2, and 3, such that the sum of all the digits is 10, we can approach this using combinatorics.

Since the sum of all digits is 10, there are several combinations possible. We can have:
1. Four 1s and three 3s: $ \binom{7}{4} $ ways to arrange these digits.
2. Two 2s, two 3s, and three 1s: $ \binom{7}{2} \times \binom{5}{2} $ ways to arrange these digits.

Calculating these combinations:
1. $ \binom{7}{4} = \frac{7!}{4!(7-4)!} = \frac{7!}{4!3!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35 $ ways.
2. $ \binom{7}{2} \times \binom{5}{2} = \frac{7!}{2!(7-2)!} \times \frac{5!}{2!(5-2)!} = \frac{7 \times 6}{2 \times 1} \times \frac{5 \times 4}{2 \times 1} = 21 \times 10 = 210 $ ways.

Therefore, the total number of seven-digit integers meeting the given criteria is $ 35 + 210 = 245 $.

The number of seven digit integer is to be formed using only 1, 2 &amp;3 such that the sum of all the digits is 10 . so how many such seven digit number is possible ?

The number of seven digit integer is to be formed using only 1, 2 &3 such that the sum of all the digits is 10 . so how many such seven digit number is possible ?