There are 10 counters to be picked from 7 different colours. Counters are all alike except colour and there are at least ten of each colour. How many arrangements have at least one counter of each colour?
Since there are 7 type of colours.So i took three cases
(1)4 alike colour counters,6 different colour counters in $\frac{7\times 10!}{4!}$
(2)3 alike colour counters of one type,2 alike colour counter of other type,5 different colour counters in $\frac{7\times 6\times 10!}{3!2!}$
(3)2 alike colour counters of one type,2 alike colour counter of second type,2 alike colour counter of third type,4 different colour counters in $\frac{7\times 6\times 5 \times 10!}{2!2!2!}$
This totals upto $\frac{103\times 7\times 10!}{24}$ but the answer given in my book is $\frac{49\times 10!}{6}$.I dont know where i am wrong.
Type 3, the choice of colors is $7\choose 3$ rather than $7\times 6\times5$