In how many ways can 7 different balls be distributed in 5 different boxes

Q: Ex 2: Suppose, that we have the following set of letters {a,b, b, c, e, e, e, f, i,i,l,n, q, q,r, s,...

A: {a,b,b,c,e,e,e,f,i,i,l,n,q,q,r,s,t,u,y} = 19 letters a 1 b 2 c 1 e 3 f 1 i 2 l 1 n ...

There are 5 different boxes and 7 different balls. All the 7 balls are to be distributed among the 5 boxes placed in a row so that any box can receive any number of balls.In how many ways can these balls be distributed so that box 2 and box 4 contain only 1 and 2 balls respectively?

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Answer: Option B

Explanation:

Solution 1: Using Formula
[Reference: Distribution of k balls into n boxes: formula 3]Initially let's find out the total number of ways in which 7 different balls can be distributed in 5 different boxes if any box can contain any number of balls and no box can be empty.Here n = 5, k = 7.Hence, as per the above formula, total number of ways= S(k,n) × n!$=\sum\limits_{i=0}^{n-1}(-1)^i~n_{C_i}(n-i)^k\\=\sum\limits_{i=0}^4(-1)^i~5_{C_i}(5-i)^7\\{\small=5_{C_0}(5)^7-5_{C_1}(4)^7+5_{C_2}(3)^7-5_{C_3}(2)^7+5_{C_4}(1)^7}\\=(5)^7-5(4)^7+10(3)^7-10(2)^7+5(1)^7\\=78125-81920+21870-1280+5\\=16800~~\cdots(1)$Now let's find out the total number of ways in which 7 different balls can be distributed in 5 different boxes if any box can contain any number of balls, no box can be empty and ball 3 and ball 5 are in the same box.For this, tie ball 3 and ball 5 and consider it as a single ball. Hence, we can consider the total number of balls as 6.

Using the same formula mentioned above, with n = 5 and k = 6,

total number of ways= S(k,n) × n!$=\sum\limits_{i=0}^{n-1}(-1)^i~n_{C_i}(n-i)^k\\=\sum\limits_{i=0}^4 (-1)^i~5_{C_i}(5-i)^6\\{\small=5_{C_0}(5)^6-5_{C_1}(4)^6+5_{C_2}(3)^6-5_{C_3}(2)^6+5_{C_4}(1)^6}\\=(5)^6-5(4)^6+10(3)^6-10(2)^6+5(1)^6\\=15625-20480+7290-640+5\\=1800~~\cdots(2)$Required number of ways =(Total Number of ways in which 7 different balls can be distributed in 5 different boxes if any box can contain any number of balls, no box can be empty)-(Total Number of ways in which 7 different balls can be distributed in 5 different boxes if any box can contain any number of balls, no box can be empty and ball 3 and ball 5 are in the same box)

$=16800-1800=15000$

Solution 2Initially let's find out the total number of ways in which 7 different balls can be distributed in 5 different boxes if any box can contain any number of balls and no box can be empty.Since no box can be left empty, there can be only two cases.

Case 1: $1, 1, 1, 1, 3$

(i.e., 3 balls are put in 1 box and 1 ball is put in each of the remaining 4 boxes.)

A box (in which 3 balls are put) can be selected in 5C1 ways.

Now, the three balls can be selected in 7C3 ways.Remaining 4 balls can be arranged in 4! ways.Hence, total number of ways

= 5C1 × 7C3 × 4! ...(A)

Case 2: $1, 1, 1, 2, 2$

(i.e., two balls are put in each of the two boxes and 1 ball is put in each of the remaining 3 boxes.)

The two boxes (in each of them, two balls are put) can be selected in 5C2 ways.

Now, two balls for the first selected box, can be selected in 7C2 ways.
Two balls for the second selected box, can be selected in 5C2 ways.Remaining 3 balls can be arranged in 3! ways.Hence, total number of ways

= 5C2 × 7C2 × 5C2 × 3! ...(B)

From (A) and (B), total number of ways

= (5C1 × 7C3 × 4!) + (5C2 × 7C2 × 5C2 × 3!)

$=(5×35×24)+(10×21×10×6)\\=4200+12600\\=16800$Now let's find out the total number of ways in which 7 different balls can be distributed in 5 different boxes if any box can contain any number of balls, no box can be empty and ball 3 and ball 5 are in the same box.For this, tie ball 3 and ball 5 and consider it as a single ball. Hence, we can consider total number of balls as 6. Now, with no box can be left empty, there can be only one case as given below.$1 , 1, 1, 1, 2$(i.e., 2 balls are put in 1 box and 1 ball is put in each of the remaining 4 boxes.)

A box (in which 2 balls are put) can be selected in 5C1 ways.

Now, the two balls can be selected in 6C2 ways.Remaining 4 balls can be arranged in 4! ways.Hence, total number of ways

= 5C1 × 6C2 × 4!

$=5×15×24\\=5×15×12×2\\=10×15×12\\=1800$Required number of ways =(Total Number of ways in which 7 different balls be distributed in 5 different boxes if any box can contain any number of balls, no box can be empty.)-(Total Number of ways in which 7 different balls be distributed in 5 different boxes if any box can contain any number of balls, no box can be empty and ball 3 and ball 5 are in the same box.)

$=16800-1800=15000$