If the number of fourteen digit numbers which can be formed by using the digits 1,2,3,4,7,7, 7, ldots .7 (digit 7 is used 10 times) such that there are atleast two identical digits between distinct digits (distinct digits are 1,2,3 and 4 ) is k ⋅ 5 ! then k is

Question

If the number of fourteen digit numbers which can be formed by using the digits 1,2,3,4,7,7, 7, ldots .7 (digit 7 is used 10 times) such that there are atleast two identical digits between distinct digits (distinct digits are 1,2,3 and 4 ) is k ⋅ 5 ! then k is (A) 3 (B) 7 (C) 14 (D) 5

Tardigrade Admin · Accepted Answer

x1 1 x2 2 x3 3 x4 4 x5 where xi denotes no. of 7 's where x2, x3, x4 ≥ 2 and x1, x5>0 Using beggar method, we get 8 C4 × 4 !=14 × 5 !

Q. If the number of fourteen digit numbers which can be formed by using the digits $1, 2, 3, 4, 7, 7$ , $7, \dots .7$ (digit 7 is used 10 times) such that there are atleast two identical digits between distinct digits (distinct digits are $1, 2, 3$ and 4 ) is $k \cdot 5$ ! then $k$ is

3

7

14

5

$x_{1} 1 x_{2} 2 x_{3} 3 x_{4} 4 x_{5}$
where $x_{i}$ denotes no. of $7^{'} s$
where $x_{2}, x_{3}, x_{4} \geq 2$
and $x_{1}, x_{5} > 0$
Using beggar method, we get
$^{8} C_{4} \times 4! = 14 \times 5!$

Q. If the number of fourteen digit numbers which can be formed by using the digits 1,2,3,4,7,7, 7,….7 (digit 7 is used 10 times) such that there are atleast two identical digits between distinct digits (distinct digits are 1,2,3 and 4 ) is k⋅5 ! then k is

3

7

14

5

x1​1x2​2x3​3x4​4x5​ where xi​ denotes no. of 7′s where x2​,x3​,x4​≥2 and x1​,x5​>0 Using beggar method, we get 8C4​×4!=14×5!

Q. If the number of fourteen digit numbers which can be formed by using the digits $1, 2, 3, 4, 7, 7$ , $7, \dots .7$ (digit 7 is used 10 times) such that there are atleast two identical digits between distinct digits (distinct digits are $1, 2, 3$ and 4 ) is $k \cdot 5$ ! then $k$ is

$x_{1} 1 x_{2} 2 x_{3} 3 x_{4} 4 x_{5}$
where $x_{i}$ denotes no. of $7^{'} s$
where $x_{2}, x_{3}, x_{4} \geq 2$
and $x_{1}, x_{5} > 0$
Using beggar method, we get
$^{8} C_{4} \times 4! = 14 \times 5!$