Question

There are $3$ books on Mathematics, $4$ on Physics and $5$ on English. How many different collections can be made such that each collection consists of:

(a) One book of each subject $\left(i\right)3968$

(b)Atleast One book of each subject $\left(ii\right)60$

(c)Atleast One book of English $\left(iii\right)3255$

Answer 1

(a) One book of each subject

Since number of ways of selecting r objects from n distinct objects is = ${}^n{C}_r$

⸫ Number of ways of selecting one book from $4$ Mathematics books = ${}^3{C}_1$ …(1)

Number of ways of selecting one book from $4$ Physics books = ${}^4{C}_1$ …(2)

Number of ways of selecting one book from $5$ English books = ${}^5{C}_1$ …(3)

Since, we need to select exactly one book from each subject simultaneously.

⸫ Number of ways of selecting exactly one book from each subject

= ${}^3{C}_1{\times }^4{C}_1{\times }^5{C}_1$ (From 1, 2 & 3)

= $(3\times 4\times 5)$ $[\because {}^n{C}_1=n]$

= $60$

(a) – (ii)

(b) At least one book of each subject.

Since number of ways of selecting atleast one object from n distinct objects.

=${}^n{C}_1{+}^n{C}_2+\dots {+}^n{C}_n=2^n-1$

⸫ Number of ways of selecting atleast one book from 3 Mathematics books

= ${}^3{C}_1{+}^3{C}_2{+}^3{C}_3$= $2^3-1=7$ …(4)

Number of ways of selecting atleast one book from $4$ Physics books

= ${}^4{C}_1{+}^4{C}_2{+}^4{C}_3{+}^4{C}_4=2^4-1$

= $15$ … (5)

Number of ways of selecting atleast one book from $5$ English books

= ${}^5{C}_1{+}^5{C}_2{+}^5{C}_3{+}^5{C}_4{+}^5{C}_5$

= $2^5-31$ … (6)

Since, we need to select at least one book from each subject simultaneously.

$\therefore$ Number of ways of selecting at least

one book from each subject

= $7\times 15\times 31$ (From 4, 5 & 6)

= $3255$

(b) – (iii)

(c) At least one book of English

Since number of ways of selecting atleast one object from n distinct objects

= ${}^n{C}_1{+}^n{C}_2+\dots {+}^n{C}_n=2^n-1$

And Number of ways of selecting any number of objects from n distinct objects

= ${}^n{C}_0{+}^n{C}_1{+}^n{C}_2+\dots {+}^n{C}_n=2^n$

$\therefore$ Number of ways of selecting none, one or more book from 3 Mathematics books

= ${}^3{C}_0{+}^3{C}_1{+}^3{C}_2{+}^3{C}_3=2^3=8$ …. (7)

Number of ways of selecting none , one or more book from 4 Physics books

= ${}^4{C}_0{+}^4{C}_1{+}^4{C}_2{+}^4{C}_3{+}^4{C}_4=2^4$

= 16 … (8)

Number of ways of selecting atleast one book from 5 English books

= ${}^5{C}_1{+}^5{C}_2{+}^5{C}_3{+}^5{C}_4{+}^5{C}_5$

=$2^5-1$

=$31$ …(9)

Since we need to select above selection simultaneously.

$\therefore$ Number of ways of selecting atleast one book from English subject

=$8\times 16\times 31$ (From 7, 8 & 9)

= $3968$

(c) – (i)