Counting starts, 1-2-3 Go....

Probability Level 3

How many seven-digit integers with digit sum 10 can be formed using the digits 1, 2, and/or 3?

Note : Not all of the digits 1, 2, 3 must appear in the integer.

77 55 88 66

2 solutions

Mehul Chaturvedi
Dec 25, 2014

First Of all the possible digits which could sum up to $10$ are

Now $1st$ no. could be arranged in $\dfrac{7!}{5!} =42$ ways

Now $2nd$ no. could be arranged in $\dfrac{7!}{4!\times 3!}=35$ ways

$\therefore$ Total ways are $35+42$ $\Rightarrow$ $\boxed{\color{#3D99F6}{\Huge 77}}$

But it is written $1,2\color{#D61F06}{\text{ and }}3$ . So answer should be 42.

Pranjal Jain - 6 years, 5 months ago

"Note: Not all of the digits 1, 2, 3 must appear in the integer." Thus, permutations of 2221111 are valid solutions.

Karl Swartz - 6 years, 1 month ago

A Former Brilliant Member
Dec 22, 2016

There are two cases.

First case: $1111123$

no. of arrangements = $\frac{7!}{5!}$ = $42$

Second case: $1111222$

no. of arrangements = $\frac{7!}{4!3!}$ = $35$

total number of arrangements = $42+35=77$