A student has $10$ blie pens, $5$ red pens, and $7$ black pens. $\newline$ $1$ ) In how many ways can he select $3$ blue pons, $2$ red pens, and $1$ black pen $\newline$ $2$ ) In how many ways can he select $5$ pens if only the choice of colors matters.

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Q. A student has

10

blie pens,

5

red pens, and

7

black pens.

\newline

1

) In how many ways can he select

3

blue pons,

2

red pens, and

1

black pen

\newline

2

) In how many ways can he select

5

pens if only the choice of colors matters.

Selecting Pens: For the first part, we need to select $3$ blue pens out of $10$ , $2$ red pens out of $5$ , and $1$ black pen out of $7$ . Use combinations: $C(n, k) = \frac{n!}{k!(n-k)!}$ . Calculate the combination for blue pens: $C(10, 3)$ .
Calculating Combinations: $\frac{10!}{3!(10-3)!} = \frac{10!}{3!7!} = \frac{10\times9\times8}{3\times2\times1} = 120$ .
Total Ways for First Part: Calculate the combination for red pens: $C(5, 2)$ .
Choosing Colors for Second Part: $\frac{5!}{2!(5-2)!} = \frac{5!}{2!3!} = \frac{5\times4}{2\times1} = 10$ .
Total Ways for Second Part: Calculate the combination for black pens: $C(7, 1)$ .
Total Ways for Second Part: Calculate the combination for black pens: $C(7, 1).7! / (1!(7-1)!) = 7! / (1!6!) = 7 / 1 = 7$ .
Total Ways for Second Part: Calculate the combination for black pens: $C(7, 1).7! / (1!(7-1)!) = 7! / (1!6!) = 7 / 1 = 7$ . Multiply the combinations for blue, red, and black pens to get the total number of ways for the first part. $\newline$ $120 \times 10 \times 7 = 8400$ .
Total Ways for Second Part: Calculate the combination for black pens: $C(7, 1).7! / (1!(7-1)!) = 7! / (1!6!) = 7 / 1 = 7$ . Multiply the combinations for blue, red, and black pens to get the total number of ways for the first part. $\newline$ $120 \times 10 \times 7 = 8400$ . For the second part, since only the choice of colors matters, we have $3$ colors to choose from and we need to select $5$ pens. $\newline$ This is a combination problem with repetition allowed: $C(n+k-1, k)$ . $\newline$ Calculate the combination for choosing $5$ pens from $3$ colors: $C(3+5-1, 5)$ .
Total Ways for Second Part: Calculate the combination for black pens: $C(7, 1).7! / (1!(7-1)!) = 7! / (1!6!) = 7 / 1 = 7$ . Multiply the combinations for blue, red, and black pens to get the total number of ways for the first part. $\newline$ $120 \times 10 \times 7 = 8400$ . For the second part, since only the choice of colors matters, we have $3$ colors to choose from and we need to select $5$ pens. $\newline$ This is a combination problem with repetition allowed: $C(n+k-1, k)$ . $\newline$ Calculate the combination for choosing $5$ pens from $3$ colors: $C(3+5-1, 5)$ . $C(7, 5) = 7! / (5!(7-5)!) = 7! / (5!2!) = (7\times6) / (2\times1) = 21$ .
Total Ways for Second Part: Calculate the combination for black pens: $C(7, 1)$ . $7! / (1!(7-1)!) = 7! / (1!6!) = 7 / 1 = 7$ . Multiply the combinations for blue, red, and black pens to get the total number of ways for the first part. $\newline$ $120 \times 10 \times 7 = 8400$ . For the second part, since only the choice of colors matters, we have $3$ colors to choose from and we need to select $5$ pens. $\newline$ This is a combination problem with repetition allowed: $C(n+k-1, k)$ . $\newline$ Calculate the combination for choosing $5$ pens from $3$ colors: $C(3+5-1, 5)$ . $C(7, 5) = 7! / (5!(7-5)!) = 7! / (5!2!) = (7\times6) / (2\times1) = 21$ . The total number of ways to select $5$ pens with only the choice of colors mattering is $7! / (1!(7-1)!) = 7! / (1!6!) = 7 / 1 = 7$ $1$ .