The number of seven digit intergers possible with sum of the digits to 11 and formed by using the digits 1, 2 and 3 only are ___________
161

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Solution

The correct option is A 161
Total possibility with sum = 11 and 7 difits

3,3,1,1,1,1,1
3,2,2,1,1,1,1
2,2,2,2,1,1,1
$3, 3, 1, 1, 1, 1, 1 \Rightarrow \frac{7!}{2! \times 5!} = 21 n u m b e r s$
$3, 2, 2, 1, 1, 1, 1 \Rightarrow \frac{7!}{2! \times 4!} = 105 n u m b e r s$
$2, 2, 2, 2, 1, 1, 1 \Rightarrow \frac{7!}{4! \times 3!} = 35 n u m b e r s$
$∴ T h e n u m b e r o f 7 d i g i t i n t e g e r s$
= 21 +105 + 35 = 161

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The number of seven digit intergers possible with sum of the digits to 11 and formed by using the digits 1, 2 and 3 only are ___________161

The correct option is A 161Total possibility with sum = 11 and 7 difits 3,3,1,1,1,1,1 3,2,2,1,1,1,1 2,2,2,2,1,1,1 3,3,1,1,1,1,1⇒7!2!×5!=21 numbers 3,2,2,1,1,1,1⇒7!2!×4!=105 numbers 2,2,2,2,1,1,1⇒7!4!×3!=35 numbers ∴ The number of 7 digit integers = 21 +105 + 35 = 161

The number of seven digit intergers possible with sum of the digits to 11 and formed by using the digits 1, 2 and 3 only are ___________
161