Question

║⊕QUESTION⊕║
Do not worry about your difficulties in Mathematics. I can assure you mine are still greater
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CLASS 11
PERMUTATIONS AND COMBINATIONS
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In how many ways can two 10 paise, two 20 paise, three 25 paise, and one 50 paise coins be distributed among 8 children so that each child gets only one coin?

Answer 1

Answer:

$\huge{\boxed{\boxed{\red{\ulcorner{\mid{\overline{\underline{\bf{solutions}}}}}\mid}}}}$

Step-by-step explanation:

the the total number of coins = 2+2+3+1

=8 coins

two coins are 10 paise, two coins are 20 paise three coins are 25 paise and one coins is 50 paise.

The required numbes are as follows:

$\frac{8 !}{2!2!3!1 !} = \frac{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{2 \times 1 \times 2 \times 1 \times 3 \times 2 \times 1 \times 1}$

After calculation we get it:

=> 8×7×6×5

=> 1680$\huge{\boxed{\boxed{\red{\bf{Answer}}}}}\mid$

Answer 2

Hence, 1680 is the answer !!