$A$ linguistic club consists of $6$ girls and $4$ boys. $A$ team of $4$ members is to be selected from this group including the selection of a leader (from among these $4$ members) for the team. If the team has to include at most one boy,the number of ways of selecting the team is
- A$140$
- B$320$
- C$76$
- D$380$
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There are $n$ white and $n$ black balls marked $1, 2, 3, \ldots, n$. The number of ways in which we can arrange these balls in a row so that neighbouring balls are of different colours is
Consider all possible permutations of the letters of the word $ENDEANOEL$. Match the Statements / Expressions in $Column I$ with the Statements / Expressions in $Column II$.
$Column I$ $Column II$ $(A)$ The number of permutations containing the word $ENDEA$ is $(p)$ $5!$ $(B)$ The number of permutations in which the letter $E$ occurs in the first and the last positions is $(q)$ $2 \times 5!$ $(C)$ The number of permutations in which none of the letters $D, L, N$ occurs in the last five positions is $(r)$ $7 \times 5!$ $(D)$ The number of permutations in which the letters $A, E, O$ occur only in odd positions is $(s)$ $21 \times 5!$
| $Column I$ | $Column II$ |
| $(A)$ The number of permutations containing the word $ENDEA$ is | $(p)$ $5!$ |
| $(B)$ The number of permutations in which the letter $E$ occurs in the first and the last positions is | $(q)$ $2 \times 5!$ |
| $(C)$ The number of permutations in which none of the letters $D, L, N$ occurs in the last five positions is | $(r)$ $7 \times 5!$ |
| $(D)$ The number of permutations in which the letters $A, E, O$ occur only in odd positions is | $(s)$ $21 \times 5!$ |
In the product $(1 + x) (1 + x + x^2) (1 + x + x^2 + x^3) \dots (1 + x + x^2 + \dots + x^{100})$,when written in ascending powers of $x$,the highest exponent of $x$ is . . . . . . .
The number of four-digit natural numbers which contain exactly two distinct digits is:
Let $A$ be a set containing $n$ elements. $A$ subset $P$ of $A$ is chosen, and the set $A$ is reconstructed by replacing the elements of $P$. $A$ subset $Q$ of $A$ is chosen again. The number of ways of choosing $P$ and $Q$ such that $Q$ contains just one element more than $P$ is
MediumWBJEE 2023
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