JEE Main · 2012 · 29 Apr 2012 · Offline · Permutations and Combinations

WhatsApp Email

Assuming the balls to be identical except for difference in colours, the number of ways in which one or more balls can be selected from 10 white, 9 green and 7 black balls is

(A)880
(B)629
(C)630
(D)879✓ Correct

Answer: (D)

Solution

For identical balls of a single colour, the number chosen can be anything from none up to all of them, giving

n + 1

possibilities, and the three colours are decided independently. With

10

9

and

7

balls the count including the empty choice is

11 \times 10 \times 8 = 880

. Removing the one case in which nothing is picked leaves

880 - 1 = 879

ways to select one or more balls.

JEE Main · 2012 · 29 Apr 2012 · Offline · Single correct option

Practise more: all JEE Main PYQs on Permutations and Combinations →

More JEE Main questions on Permutations and Combinations

The number of strictly increasing functions f from the set ${1, 2, 3, 4, 5, 6}$ to the set $(1, 2, 3, \dots, 9)$ such that $f (i) \neq = i$ for $1 \leq i \leq 6$ , is equal to :
JEE Main 2026View solution →
Let $S = {(m, n) : m, n \in {1, 2, 3, \dots .., 50}}$ . If the number of elements ( $m, n$ ) in S such that $6^{m} + 9^{n}$ is a multiple of 5 is p and the number of elements …
JEE Main 2026View solution →
The largest $n \in N$ , for which $7^{n}$ divides 101!, is:
JEE Main 2026View solution →
Let ABC be a triangle. Consider four points $p_{1}, p_{2}$ , $p_{3}, p_{4}$ on the side AB , five points $p_{5}, p_{6}, p_{7}, p_{8}, p_{9}$ on the side BC and four points …
JEE Main 2026View solution →
Let S be the set of the first 11 natural numbers. Then the number of elements in $A = {B \subseteq S : n (B) \geq 2$ and the product of all elements of B is even $}$ is
JEE Main 2026View solution →
The number of 4-letter words, with or without meaning, which can be formed using the letters PQRPQRSTUVP, is ______.
JEE Main 2026View solution →