$A$ debate club consists of $6$ girls and $4$ boys. $A$ team of $4$ members is to be selected from this club,including the selection of a captain (from among these $4$ members) for the team. If the team has to include at most one boy,then the number of ways of selecting the team is
- A$380$
- B$320$
- C$260$
- D$95$
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If $\alpha = ^mC_2$,then $^\alpha C_2$ is equal to
Medium
View SolutionConsider 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 | $(s)$ $21 \times 5!$ |
The number of all $3$-digit numbers $abc$ (in base $10$) for which $(a \times b \times c) + (a \times b) + (b \times c) + (c \times a) + a + b + c = 29$ is:
The number of integers greater than $6000$ that can be formed using the digits $3, 5, 6, 7,$ and $8$ without repetition is:
DifficultJEE MAIN 2014
View SolutionLet $n$ be the number of ways in which $5$ boys and $5$ girls can stand in a queue such that all the girls stand consecutively. Let $m$ be the number of ways in which $5$ boys and $5$ girls can stand in a queue such that exactly four girls stand consecutively. Then the value of $\frac{m}{n}$ is
MediumIIT 2015
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