The number of ways in which 18 identical coins can be distributed among five children A, B, C, D and E so that everyone gets atleast one coin but not more than 8 coins, is

Question

The number of ways in which 18 identical coins can be distributed among five children A, B, C, D and E so that everyone gets atleast one coin but not more than 8 coins, is (A) 2380 (B) 2150 (C) 1750 (D) 1650

Tardigrade Admin · Accepted Answer

Giving one to each we have 13 coins to be distributed in 5 children.
OOOOOOOOOOOOOOØØØØ
using beggar

^{17} C_{4}

ways
Now, give 8 more to

A

or

B

or

C

or

D

or

E

we have number of ways to be rejected

ООООО ØØØ Ø=5.^{9} C_{4}

Required ways

=^{17} C_{4} - 5 \cdot^{9} C_{4} = 2380 - 630 = 1750

ways

Q. The number of ways in which 18 identical coins can be distributed among five children A, B, C, D and E so that everyone gets atleast one coin but not more than 8 coins, is

2380

2150

1750

1650

Giving one to each we have 13 coins to be distributed in 5 children. OOOOOOOOOOOOOOØØØØ using beggar 17C4​ ways Now, give 8 more to A or B or C or D or E we have number of ways to be rejected ОООООØØØ Ø=5. 9C4​ Required ways =17C4​−5⋅9C4​=2380−630=1750 ways