The number of ways of distributing 50 identical things among 8 persons in such a way that three of them get 8 things each, two of them get 7 things each, and remaining 3 get 4 things each, is equal to

Question

The number of ways of distributing 50 identical things among 8 persons in such a way that three of them get 8 things each, two of them get 7 things each, and remaining 3 get 4 things each, is equal to (A) ((50!)(8!)/(8!)3(3!)2(7!)2 (4!)3(2!)) (B) ((50!)(8!)/(8!)3(7!)2 (4!)3) (C) ((50!)/(8!)3(7!)2 (4!)3) (D) ((8!)/(3!)2 (2!))

Tardigrade Admin · Accepted Answer

Number of ways of dividing 8 persons in three groups, first having 3 persons, second having 2 persons and third having 3 persons = (8!/3! 2! 3!). Since all the 50 things are identical, so, required number = ((8!)/(3!)2⋅(2!))

Q. The number of ways of distributing $50$ identical things among $8$ persons in such a way that three of them get $8$ things each, two of them get $7$ things each, and remaining $3$ get $4$ things each, is equal to

$\frac{( 50 ! ) ( 8 ! )}{( 8 ! ) ^{3} ( 3 ! ) ^{2} ( 7 ! ) ^{2} ( 4 ! ) ^{3} ( 2 ! )}$

$\frac{( 50 ! ) ( 8 ! )}{( 8 ! ) ^{3} ( 7 ! ) ^{2} ( 4 ! ) ^{3}}$

$\frac{( 50 ! )}{( 8 ! ) ^{3} ( 7 ! ) ^{2} ( 4 ! ) ^{3}}$

$\frac{( 8 ! )}{( 3 ! ) ^{2} ( 2 ! )}$

Number of ways of dividing $8$ persons in three groups,
first having $3$ persons,
second having $2$ persons
and third having $3$ persons = $\frac{8 !}{3 ! 2 ! 3 !}$ .
Since all the $50$ things are identical,
so, required number = $\frac{( 8 ! )}{( 3 ! ) ^{2} \cdot ( 2 ! )}$

Q. The number of ways of distributing 50 identical things among 8 persons in such a way that three of them get 8 things each, two of them get 7 things each, and remaining 3 get 4 things each, is equal to

(8!)3(3!)2(7!)2(4!)3(2!)(50!)(8!)​

(8!)3(7!)2(4!)3(50!)(8!)​

(8!)3(7!)2(4!)3(50!)​

(3!)2(2!)(8!)​